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<?xml version="1.0" encoding="utf-8"?>
<feed xmlns="http://www.w3.org/2005/Atom">
<title>Andy's Blog</title>
<subtitle>Quick notes</subtitle>
<link href="/atom.xml" rel="self"/>
<link href="https://andy.city/"/>
<updated>2022-10-07T10:10:56.000Z</updated>
<id>https://andy.city/</id>
<author>
<name>Andy</name>
</author>
<generator uri="http://hexo.io/">Hexo</generator>
<entry>
<title>LeetCode-347.前K个高频元素</title>
<link href="https://andy.city/archives/df8f.html"/>
<id>https://andy.city/archives/df8f.html</id>
<published>2022-10-07T10:10:56.000Z</published>
<updated>2022-10-07T10:10:56.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/top-k-frequent-elements/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/0347.前K个高频元素.html" target="_blank" rel="noopener">代码随想录</a></li></ul><iframe class="iframe_video" src="//player.bilibili.com/player.html?aid=514643371&bvid=BV1Xg41167Lz&cid=808260290&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen></iframe><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><ul><li>先统计每个元素出现次数，再使用优先队列，最后输出</li></ul><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/0347.前K个高频元素.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>为什么使用优先队列呢？因为直接排序时间复杂度一般是 O(n log n)</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><ul><li>使用小顶堆而不是大顶堆，因为：<ul><li>如果用大顶堆，时间复杂度是 O(n log n)（高频元素在栈顶，那么堆就需要维护所有节点，最大是n）</li><li>如果用小顶堆，时间复杂度是 O(n log k)（低频元素在栈顶被出栈，栈只需要维护k个节点，最终留下的k个元素就是前k个高频元素），明显时间复杂度要优一些</li></ul></li></ul><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">topKFrequent</span>(<span class="params">nums: <span class="built_in">number</span>[], k: <span class="built_in">number</span></span>): <span class="title">number</span>[] </span>{</span><br><span class="line"> <span class="keyword">const</span> data: { [k: <span class="built_in">number</span>]: <span class="built_in">number</span> } = nums.reduce(<span class="function">(<span class="params">acc, cur</span>) =></span> (acc[cur] = (acc[cur] || <span class="number">0</span>) + <span class="number">1</span>, acc), {})</span><br><span class="line"> <span class="keyword">const</span> queue = <span class="keyword">new</span> MinPriorityQueue({ priority: <span class="function">(<span class="params">item: [<span class="built_in">number</span>, <span class="built_in">number</span>]</span>) =></span> item[<span class="number">1</span>] })</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">const</span> item of <span class="built_in">Object</span>.entries(data)) {</span><br><span class="line"> queue.enqueue(item)</span><br><span class="line"> <span class="keyword">if</span> (queue.size() > k) queue.dequeue()</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">const</span> res: <span class="built_in">number</span>[] = []</span><br><span class="line"> <span class="keyword">while</span> (queue.size()) res.push(queue.dequeue().element[<span class="number">0</span>])</span><br><span class="line"> <span class="keyword">return</span> res</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n log k)<br>空间复杂度：O(k)</p><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="队列" scheme="https://andy.city/tags/%E9%98%9F%E5%88%97/"/>
<category term="优先队列" scheme="https://andy.city/tags/%E4%BC%98%E5%85%88%E9%98%9F%E5%88%97/"/>
<category term="堆" scheme="https://andy.city/tags/%E5%A0%86/"/>
<category term="小顶堆" scheme="https://andy.city/tags/%E5%B0%8F%E9%A1%B6%E5%A0%86/"/>
</entry>
<entry>
<title>LeetCode-239.滑动窗口最大值</title>
<link href="https://andy.city/archives/6798.html"/>
<id>https://andy.city/archives/6798.html</id>
<published>2022-10-07T08:12:13.000Z</published>
<updated>2022-10-07T08:12:13.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/sliding-window-maximum/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/0239.滑动窗口最大值.html" target="_blank" rel="noopener">代码随想录</a></li></ul><iframe class="iframe_video" src="//player.bilibili.com/player.html?aid=729559279&bvid=BV1XS4y1p7qj&cid=806301877&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen></iframe><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><ul><li>暴力解法时间复杂度是 O(n^2)，而用单调队列只有 O(n)<ul><li>使用双端队列实现单调递减队列，队顶始终是最大值</li><li>每次滑动窗口时，均有元素入队和出对，然后队顶元素就是当前窗口的最大值</li></ul></li></ul><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/0239.滑动窗口最大值.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>将单调队列的代码抽离封装好，主流程十分简单已读</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><ul><li>实现单调队列时入队逻辑</li></ul><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">maxSlidingWindow</span>(<span class="params">nums: <span class="built_in">number</span>[], k: <span class="built_in">number</span></span>): <span class="title">number</span>[] </span>{</span><br><span class="line"> <span class="keyword">const</span> res: <span class="built_in">number</span>[] = [],</span><br><span class="line"> queue = <span class="keyword">new</span> MonoQueue<<span class="built_in">number</span>>()</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">0</span>; i < nums.length; i++) {</span><br><span class="line"> queue.enqueue(nums[i])</span><br><span class="line"> <span class="keyword">if</span> (i >= k) queue.dequeue(nums[i - k])</span><br><span class="line"> <span class="keyword">if</span> (i >= k - <span class="number">1</span>) res.push(queue.top())</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> res</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="comment">/** 单调递减队列 */</span></span><br><span class="line"><span class="keyword">class</span> MonoQueue<T> {</span><br><span class="line"> queue: T[]</span><br><span class="line"> <span class="keyword">constructor</span>(<span class="params"></span>) {</span><br><span class="line"> <span class="keyword">this</span>.queue = []</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> enqueue(value: T): <span class="built_in">void</span> {</span><br><span class="line"> <span class="comment">// 比value小的都剔除</span></span><br><span class="line"> <span class="keyword">while</span> (<span class="keyword">this</span>.queue.length && value > <span class="keyword">this</span>.queue[<span class="keyword">this</span>.queue.length - <span class="number">1</span>]) <span class="keyword">this</span>.queue.pop()</span><br><span class="line"> <span class="keyword">this</span>.queue.push(value)</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> dequeue(value: T): <span class="built_in">void</span> {</span><br><span class="line"> <span class="comment">// value可能在前面元素push的时候已经被剔除了</span></span><br><span class="line"> <span class="keyword">if</span> (value === <span class="keyword">this</span>.top()) <span class="keyword">this</span>.queue.shift()</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> top(): T {</span><br><span class="line"> <span class="keyword">return</span> <span class="keyword">this</span>.queue[<span class="number">0</span>]</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n)<br>空间复杂度：O(n)</p><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2><ul><li>学习了单调队列</li></ul>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="队列" scheme="https://andy.city/tags/%E9%98%9F%E5%88%97/"/>
<category term="单调队列" scheme="https://andy.city/tags/%E5%8D%95%E8%B0%83%E9%98%9F%E5%88%97/"/>
</entry>
<entry>
<title>LeetCode-150.逆波兰表达式求值</title>
<link href="https://andy.city/archives/f533.html"/>
<id>https://andy.city/archives/f533.html</id>
<published>2022-10-07T07:26:57.000Z</published>
<updated>2022-10-07T07:26:57.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/evaluate-reverse-polish-notation/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/0150.逆波兰表达式求值.html" target="_blank" rel="noopener">代码随想录</a></li></ul><iframe class="iframe_video" src="//player.bilibili.com/player.html?aid=387057277&bvid=BV1kd4y1o7on&cid=804365306&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen></iframe><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><ul><li>遍历tokens：<ul><li>如果遇到数字入栈</li><li>如果遇到运算符，连续出栈两个数字进行运算，再将运算结果入栈用于后续计算</li></ul></li></ul><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/0150.逆波兰表达式求值.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>get了二进制取整小tip</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><ul><li>出栈两次得到的两个数字顺序与运算顺序时相反的</li><li>除法取整时需要处理一下</li></ul><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">evalRPN</span>(<span class="params">tokens: <span class="built_in">string</span>[]</span>): <span class="title">number</span> </span>{</span><br><span class="line"> <span class="keyword">const</span> stack: <span class="built_in">number</span>[] = []</span><br><span class="line"> <span class="comment">// 注意调用时传入的两个数字是依次出栈得到的，顺序是反的</span></span><br><span class="line"> <span class="keyword">const</span> map = {</span><br><span class="line"> <span class="string">'+'</span>: <span class="function">(<span class="params">a: <span class="built_in">number</span>, b: <span class="built_in">number</span></span>) =></span> b + a,</span><br><span class="line"> <span class="string">'-'</span>: <span class="function">(<span class="params">a: <span class="built_in">number</span>, b: <span class="built_in">number</span></span>) =></span> b - a,</span><br><span class="line"> <span class="string">'*'</span>: <span class="function">(<span class="params">a: <span class="built_in">number</span>, b: <span class="built_in">number</span></span>) =></span> b * a,</span><br><span class="line"> <span class="string">'/'</span>: <span class="function">(<span class="params">a: <span class="built_in">number</span>, b: <span class="built_in">number</span></span>) =></span> (b / a) | <span class="number">0</span>,</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">const</span> c of tokens) {</span><br><span class="line"> <span class="keyword">if</span> (c <span class="keyword">in</span> map) stack.push(map[c](stack.pop(), stack.pop()))</span><br><span class="line"> <span class="keyword">else</span> stack.push(+c)</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> stack[<span class="number">0</span>]</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n)<br>空间复杂度：O(n)</p><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="栈" scheme="https://andy.city/tags/%E6%A0%88/"/>
</entry>
<entry>
<title>LeetCode-1047.删除字符串中的所有相邻重复项</title>
<link href="https://andy.city/archives/e442.html"/>
<id>https://andy.city/archives/e442.html</id>
<published>2022-10-07T07:10:49.000Z</published>
<updated>2022-10-07T07:10:49.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/remove-all-adjacent-duplicates-in-string/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/1047.删除字符串中的所有相邻重复项.html" target="_blank" rel="noopener">代码随想录</a></li></ul><iframe class="iframe_video" src="//player.bilibili.com/player.html?aid=216931465&bvid=BV12a411P7mw&cid=801359937&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen></iframe><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><ul><li>遍历字符串：<ul><li>如果当前元素与栈顶元素相同，则进行出栈</li><li>如果不相同，则入栈</li></ul></li><li>最后栈里的所有元素就是结果</li></ul><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/1047.删除字符串中的所有相邻重复项.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>思路一致</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">removeDuplicates</span>(<span class="params">s: <span class="built_in">string</span></span>): <span class="title">string</span> </span>{</span><br><span class="line"> <span class="keyword">const</span> stack: <span class="built_in">string</span>[] = []</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">const</span> c of s) {</span><br><span class="line"> <span class="keyword">if</span> (stack.length && stack[stack.length - <span class="number">1</span>] === c) stack.pop()</span><br><span class="line"> <span class="keyword">else</span> stack.push(c)</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> stack.join(<span class="string">''</span>)</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n)<br>空间复杂度：O(n)</p><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="栈" scheme="https://andy.city/tags/%E6%A0%88/"/>
</entry>
<entry>
<title>LeetCode-20.有效的括号</title>
<link href="https://andy.city/archives/c7da.html"/>
<id>https://andy.city/archives/c7da.html</id>
<published>2022-10-07T06:00:25.000Z</published>
<updated>2022-10-07T06:00:25.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/valid-parentheses/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/0020.有效的括号.html" target="_blank" rel="noopener">代码随想录</a></li></ul><iframe class="iframe_video" src="//player.bilibili.com/player.html?aid=301823875&bvid=BV1AF411w78g&cid=799270643&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen></iframe><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><ul><li>遍历字符串<ul><li>遇到左括号->入栈;</li><li>遇到右括号->出栈 并判断出栈元素是否与当前的遍历到的右括号成对，不成对则终止</li></ul></li></ul><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/0020.有效的括号.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>使用字典来简化判断逻辑</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">isValid</span>(<span class="params">s: <span class="built_in">string</span></span>): <span class="title">boolean</span> </span>{</span><br><span class="line"> <span class="keyword">if</span> (s.length % <span class="number">2</span> !== <span class="number">0</span>) <span class="keyword">return</span> <span class="literal">false</span></span><br><span class="line"> <span class="keyword">const</span> stack: <span class="built_in">string</span>[] = []</span><br><span class="line"> <span class="keyword">const</span> map = {</span><br><span class="line"> <span class="string">'('</span>: <span class="string">')'</span>,</span><br><span class="line"> <span class="string">'['</span>: <span class="string">']'</span>,</span><br><span class="line"> <span class="string">'{'</span>: <span class="string">'}'</span></span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">const</span> c of s) {</span><br><span class="line"> <span class="keyword">if</span> (c <span class="keyword">in</span> map) stack.push(c)</span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span> (map[stack.pop()] !== c) <span class="keyword">return</span> <span class="literal">false</span></span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> !stack.length</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n)<br>空间复杂度：O(n)</p><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2><ul><li><strong>数据结构与算法的应用往往隐藏在我们看不到的地方</strong></li></ul>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="栈" scheme="https://andy.city/tags/%E6%A0%88/"/>
</entry>
<entry>
<title>LeetCode-225.用队列实现栈</title>
<link href="https://andy.city/archives/fd25.html"/>
<id>https://andy.city/archives/fd25.html</id>
<published>2022-10-06T15:23:15.000Z</published>
<updated>2022-10-06T15:23:15.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/implement-stack-using-queues/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/0225.用队列实现栈.html" target="_blank" rel="noopener">代码随想录</a></li></ul><iframe class="iframe_video" src="//player.bilibili.com/player.html?aid=344329109&bvid=BV1Fd4y1K7sm&cid=797278452&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen></iframe><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><ul><li>两个队列互相倒腾：<ul><li>pop时将队列1中的元素队列并入队到队列2，剩最后一个即pop的目标元素</li><li>如果pop时队列为空，则将队列2与队列1内容互换</li></ul></li></ul><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/0225.用队列实现栈.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>学会了只用一个队列也能实现栈</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><ul><li><p>一个队列实现栈</p><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">/**</span></span><br><span class="line"><span class="comment">* Your MyStack object will be instantiated and called as such:</span></span><br><span class="line"><span class="comment">* var obj = new MyStack()</span></span><br><span class="line"><span class="comment">* obj.push(x)</span></span><br><span class="line"><span class="comment">* var param_2 = obj.pop()</span></span><br><span class="line"><span class="comment">* var param_3 = obj.top()</span></span><br><span class="line"><span class="comment">* var param_4 = obj.empty()</span></span><br><span class="line"><span class="comment">*/</span></span><br><span class="line"></span><br><span class="line"><span class="keyword">class</span> MyStack {</span><br><span class="line"> queue: <span class="built_in">number</span>[]</span><br><span class="line"> <span class="keyword">constructor</span>(<span class="params"></span>) {</span><br><span class="line"> <span class="keyword">this</span>.queue = []</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> push(x: <span class="built_in">number</span>): <span class="built_in">void</span> {</span><br><span class="line"> <span class="keyword">this</span>.queue.push(x)</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> pop(): <span class="built_in">number</span> {</span><br><span class="line"> <span class="keyword">let</span> len = <span class="keyword">this</span>.queue.length</span><br><span class="line"> <span class="keyword">while</span> (--len) <span class="keyword">this</span>.queue.push(<span class="keyword">this</span>.queue.shift())</span><br><span class="line"> <span class="keyword">return</span> <span class="keyword">this</span>.queue.shift()</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> top(): <span class="built_in">number</span> {</span><br><span class="line"> <span class="keyword">const</span> top = <span class="keyword">this</span>.pop()</span><br><span class="line"> <span class="keyword">this</span>.queue.push(top)</span><br><span class="line"> <span class="keyword">return</span> top</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> empty(): <span class="built_in">boolean</span> {</span><br><span class="line"> <span class="keyword">return</span> !<span class="keyword">this</span>.queue.length</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure></li><li><p>两个队列实现栈</p><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">/**</span></span><br><span class="line"><span class="comment">* Your MyStack object will be instantiated and called as such:</span></span><br><span class="line"><span class="comment">* var obj = new MyStack()</span></span><br><span class="line"><span class="comment">* obj.push(x)</span></span><br><span class="line"><span class="comment">* var param_2 = obj.pop()</span></span><br><span class="line"><span class="comment">* var param_3 = obj.top()</span></span><br><span class="line"><span class="comment">* var param_4 = obj.empty()</span></span><br><span class="line"><span class="comment">*/</span></span><br><span class="line"></span><br><span class="line"><span class="keyword">class</span> MyStack {</span><br><span class="line"> queue1: <span class="built_in">number</span>[]</span><br><span class="line"> queue2: <span class="built_in">number</span>[]</span><br><span class="line"> <span class="keyword">constructor</span>(<span class="params"></span>) {</span><br><span class="line"> <span class="keyword">this</span>.queue1 = []</span><br><span class="line"> <span class="keyword">this</span>.queue2 = []</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> push(x: <span class="built_in">number</span>): <span class="built_in">void</span> {</span><br><span class="line"> <span class="keyword">this</span>.queue1.push(x)</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> pop(): <span class="built_in">number</span> {</span><br><span class="line"> <span class="keyword">if</span> (!<span class="keyword">this</span>.queue1.length) [<span class="keyword">this</span>.queue1, <span class="keyword">this</span>.queue2] = [<span class="keyword">this</span>.queue2, <span class="keyword">this</span>.queue1]</span><br><span class="line"> <span class="keyword">while</span> (<span class="keyword">this</span>.queue1.length > <span class="number">1</span>) <span class="keyword">this</span>.queue2.push(<span class="keyword">this</span>.queue1.shift())</span><br><span class="line"> <span class="keyword">return</span> <span class="keyword">this</span>.queue1.shift()</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> top(): <span class="built_in">number</span> {</span><br><span class="line"> <span class="keyword">const</span> top = <span class="keyword">this</span>.pop()</span><br><span class="line"> <span class="keyword">this</span>.queue1.push(top)</span><br><span class="line"> <span class="keyword">return</span> top</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> empty(): <span class="built_in">boolean</span> {</span><br><span class="line"> <span class="keyword">return</span> !<span class="keyword">this</span>.queue1.length && !<span class="keyword">this</span>.queue2.length</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure></li></ul><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2><ul><li>学会了极致的的代码复用。本题中<code>top</code>函数就完全复用了<code>pop</code>函数逻辑</li></ul>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="栈" scheme="https://andy.city/tags/%E6%A0%88/"/>
<category term="队列" scheme="https://andy.city/tags/%E9%98%9F%E5%88%97/"/>
</entry>
<entry>
<title>LeetCode-232.用栈实现队列</title>
<link href="https://andy.city/archives/5c9c.html"/>
<id>https://andy.city/archives/5c9c.html</id>
<published>2022-10-06T15:03:13.000Z</published>
<updated>2022-10-06T15:03:13.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/implement-queue-using-stacks/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/0232.用栈实现队列.html" target="_blank" rel="noopener">代码随想录</a></li></ul><iframe class="iframe_video" src="//player.bilibili.com/player.html?aid=644140871&bvid=BV1nY4y1w7VC&cid=794448489&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen></iframe><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><ul><li>利用两个栈来模拟</li></ul><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/0232.用栈实现队列.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>重点是利用第二个栈来模拟出队列行为</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><ul><li>注意<code>popStack</code>中已有元素时，不能再向其中push了，否则就乱了（因为<code>popStack</code>的栈顶元素实际是模拟队列顶：即将要出队列的元素）</li></ul><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">/**</span></span><br><span class="line"><span class="comment"> * Your MyQueue object will be instantiated and called as such:</span></span><br><span class="line"><span class="comment"> * var obj = new MyQueue()</span></span><br><span class="line"><span class="comment"> * obj.push(x)</span></span><br><span class="line"><span class="comment"> * var param_2 = obj.pop()</span></span><br><span class="line"><span class="comment"> * var param_3 = obj.peek()</span></span><br><span class="line"><span class="comment"> * var param_4 = obj.empty()</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"></span><br><span class="line"><span class="keyword">class</span> MyQueue {</span><br><span class="line"> pushStack: <span class="built_in">number</span>[]</span><br><span class="line"> popStack: <span class="built_in">number</span>[]</span><br><span class="line"> <span class="keyword">constructor</span>(<span class="params"></span>) {</span><br><span class="line"> <span class="keyword">this</span>.pushStack = []</span><br><span class="line"> <span class="keyword">this</span>.popStack = []</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> push(x: <span class="built_in">number</span>): <span class="built_in">void</span> {</span><br><span class="line"> <span class="keyword">this</span>.pushStack.push(x)</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> pop(): <span class="built_in">number</span> {</span><br><span class="line"> <span class="keyword">if</span> (!<span class="keyword">this</span>.popStack.length) {</span><br><span class="line"> <span class="keyword">while</span> (<span class="keyword">this</span>.pushStack.length) <span class="keyword">this</span>.popStack.push(<span class="keyword">this</span>.pushStack.pop())</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="keyword">this</span>.popStack.pop()</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> peek(): <span class="built_in">number</span> {</span><br><span class="line"> <span class="keyword">const</span> top = <span class="keyword">this</span>.pop()</span><br><span class="line"> <span class="keyword">this</span>.popStack.push(top)</span><br><span class="line"> <span class="keyword">return</span> top</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> empty(): <span class="built_in">boolean</span> {</span><br><span class="line"> <span class="keyword">return</span> !(<span class="keyword">this</span>.pushStack.length || <span class="keyword">this</span>.popStack.length)</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2><ul><li><strong>一定要懂得复用，功能相近的函数要抽象出来，不要大量的复制粘贴，很容易出问题！（踩过坑的人自然懂）</strong></li></ul>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="栈" scheme="https://andy.city/tags/%E6%A0%88/"/>
<category term="队列" scheme="https://andy.city/tags/%E9%98%9F%E5%88%97/"/>
</entry>
<entry>
<title>LeetCode-459.重复的子字符串</title>
<link href="https://andy.city/archives/e0bf.html"/>
<id>https://andy.city/archives/e0bf.html</id>
<published>2022-10-06T14:23:56.000Z</published>
<updated>2022-10-06T14:23:56.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/repeated-substring-pattern/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/0459.重复的子字符串.html" target="_blank" rel="noopener">代码随想录</a></li></ul><iframe class="iframe_video" src="//player.bilibili.com/player.html?aid=514040419&bvid=BV1cg41127fw&cid=792550174&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen></iframe><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><ul><li>想到了用KMP，但是没想明白前缀表的用处</li></ul><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/0459.重复的子字符串.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>get了移动字符串匹配解法</li><li>利用前缀表 通过数学推导出最小重复子串：除去最大公共前后缀的剩余部分就是最小重复子串</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><ul><li>KMP求前缀表</li><li>前缀表的作用</li></ul><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><ul><li><p>移动字符串匹配</p><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">repeatedSubstringPattern</span>(<span class="params">s: <span class="built_in">string</span></span>): <span class="title">boolean</span> </span>{</span><br><span class="line"> <span class="keyword">return</span> (s + s).slice(<span class="number">1</span>, s.length * <span class="number">2</span> - <span class="number">1</span>).includes(s)</span><br><span class="line">}</span><br></pre></td></tr></table></figure></li><li><p>KMP</p><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">repeatedSubstringPattern</span>(<span class="params">s: <span class="built_in">string</span></span>): <span class="title">boolean</span> </span>{</span><br><span class="line"> <span class="keyword">const</span> prefix = getPrefix(s),</span><br><span class="line"> lastPrefix = prefix[prefix.length - <span class="number">1</span>]</span><br><span class="line"> <span class="keyword">return</span> lastPrefix !== <span class="number">0</span> && (s.length % (s.length - lastPrefix)) === <span class="number">0</span></span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">getPrefix</span>(<span class="params">s: <span class="built_in">string</span></span>): <span class="title">number</span>[] </span>{</span><br><span class="line"> <span class="keyword">let</span> j = <span class="number">0</span></span><br><span class="line"> <span class="keyword">const</span> prefix: <span class="built_in">number</span>[] = [j]</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">1</span>; i < s.length; i++) {</span><br><span class="line"> <span class="keyword">while</span> (j > <span class="number">0</span> && s[i] !== s[j]) j = prefix[j - <span class="number">1</span>]</span><br><span class="line"> <span class="keyword">if</span> (s[i] === s[j]) j++</span><br><span class="line"> prefix[i] = j</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> prefix</span><br><span class="line">}</span><br></pre></td></tr></table></figure></li></ul><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2><ul><li>get了利用KMP求最小重复子串的方法</li></ul>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="字符串" scheme="https://andy.city/tags/%E5%AD%97%E7%AC%A6%E4%B8%B2/"/>
<category term="KMP" scheme="https://andy.city/tags/KMP/"/>
</entry>
<entry>
<title>LeetCode-28.找出字符串中第一个匹配项的下标</title>
<link href="https://andy.city/archives/c765.html"/>
<id>https://andy.city/archives/c765.html</id>
<published>2022-10-06T13:03:58.000Z</published>
<updated>2022-10-06T13:03:58.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/find-the-index-of-the-first-occurrence-in-a-string/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/0028.实现strStr.html" target="_blank" rel="noopener">代码随想录</a></li></ul><iframe class="iframe_video" src="//player.bilibili.com/player.html?aid=712030093&bvid=BV1PD4y1o7nd&cid=493095580&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen></iframe><br><iframe class="iframe_video" src="//player.bilibili.com/player.html?aid=457239548&bvid=BV1M5411j7Xx&cid=496522111&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen></iframe><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><ul><li>解决字符串匹配问题的经典算法：KMP</li></ul><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/0028.实现strStr.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><ul><li>学习理解KMP算法<ul><li>理解KMP算法的原理，它是如何加速字符串匹配的</li><li>生成前缀表</li><li>利用前缀表写出匹配过程</li></ul></li></ul><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">strStr</span>(<span class="params">haystack: <span class="built_in">string</span>, needle: <span class="built_in">string</span></span>): <span class="title">number</span> </span>{</span><br><span class="line"> <span class="keyword">const</span> prefix = getPrefix(needle)</span><br><span class="line"> <span class="keyword">let</span> j = <span class="number">0</span></span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">0</span>; i < haystack.length; i++) {</span><br><span class="line"> <span class="keyword">while</span> (j > <span class="number">0</span> && haystack[i] !== needle[j]) j = prefix[j - <span class="number">1</span>]</span><br><span class="line"> <span class="keyword">if</span> (haystack[i] === needle[j]) {</span><br><span class="line"> <span class="keyword">if</span> (j === needle.length - <span class="number">1</span>) <span class="keyword">return</span> i - j</span><br><span class="line"> j++</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="number">-1</span></span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="comment">/** 求KMP最长公共前后缀 */</span></span><br><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">getPrefix</span>(<span class="params">str: <span class="built_in">string</span></span>): <span class="title">number</span>[] </span>{</span><br><span class="line"> <span class="keyword">let</span> j = <span class="number">0</span></span><br><span class="line"> <span class="keyword">const</span> prefix: <span class="built_in">number</span>[] = [j]</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">1</span>; i < str.length; i++) {</span><br><span class="line"> <span class="keyword">while</span> (j > <span class="number">0</span> && str[i] !== str[j]) j = prefix[j - <span class="number">1</span>]</span><br><span class="line"> <span class="keyword">if</span> (str[i] === str[j]) j++</span><br><span class="line"> prefix[i] = j</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> prefix</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(m+n)<br>空间复杂度：O(n)</p><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2><ul><li>系统学习了KMP算法</li></ul>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="字符串" scheme="https://andy.city/tags/%E5%AD%97%E7%AC%A6%E4%B8%B2/"/>
<category term="KMP" scheme="https://andy.city/tags/KMP/"/>
</entry>
<entry>
<title>LeetCode-剑指Offer-58-II.左旋转字符串</title>
<link href="https://andy.city/archives/8a8.html"/>
<id>https://andy.city/archives/8a8.html</id>
<published>2022-10-04T13:40:39.000Z</published>
<updated>2022-10-04T13:40:39.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/zuo-xuan-zhuan-zi-fu-chuan-lcof/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/剑指Offer58-II.左旋转字符串.html" target="_blank" rel="noopener">代码随想录</a></li></ul><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><ol><li>便历一遍，前面部分动态填充到数组尾部</li><li>后面部分原地修改数组</li><li>截取长度</li></ol><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/剑指Offer58-II.左旋转字符串.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>学习到了<code>局部反转+整体反转</code>的解法</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><ul><li><p>局部反转+整体反转</p><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">reverseLeftWords</span>(<span class="params">s: <span class="built_in">string</span>, n: <span class="built_in">number</span></span>): <span class="title">string</span> </span>{</span><br><span class="line"> <span class="keyword">const</span> strArr = s.split(<span class="string">''</span>),</span><br><span class="line"> len = s.length</span><br><span class="line"> reverse(strArr, <span class="number">0</span>, n - <span class="number">1</span>)</span><br><span class="line"> reverse(strArr, n, len - <span class="number">1</span>)</span><br><span class="line"> reverse(strArr, <span class="number">0</span>, len - <span class="number">1</span>)</span><br><span class="line"> <span class="keyword">return</span> strArr.join(<span class="string">''</span>)</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="comment">/** 反转数组元素 */</span></span><br><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">reverse</span>(<span class="params">strArr: <span class="built_in">string</span>[], start: <span class="built_in">number</span>, end: <span class="built_in">number</span></span>) </span>{</span><br><span class="line"> <span class="keyword">while</span> (start < end) {</span><br><span class="line"> [strArr[start], strArr[end]] = [strArr[end], strArr[start]]</span><br><span class="line"> start++</span><br><span class="line"> end--</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n)<br>空间复杂度：O(n)</p></li><li><p>填充</p><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">reverseLeftWords</span>(<span class="params">s: <span class="built_in">string</span>, n: <span class="built_in">number</span></span>): <span class="title">string</span> </span>{</span><br><span class="line"> <span class="keyword">const</span> strArr = s.split(<span class="string">''</span>),</span><br><span class="line"> len = s.length</span><br><span class="line"> <span class="keyword">let</span> slow = <span class="number">0</span>,</span><br><span class="line"> fast = <span class="number">0</span></span><br><span class="line"> <span class="keyword">while</span> (fast < len + n) {</span><br><span class="line"> <span class="keyword">if</span> (fast < n) strArr.push(strArr[fast++])</span><br><span class="line"> <span class="keyword">else</span> strArr[slow++] = strArr[fast++]</span><br><span class="line"> }</span><br><span class="line"> strArr.length = len</span><br><span class="line"> <span class="keyword">return</span> strArr.join(<span class="string">''</span>)</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n)<br>空间复杂度：O(n)</p></li></ul><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2><ul><li>字符串反转有时候 局部反转、整体反转 能更巧妙的解决问题</li></ul>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="双指针" scheme="https://andy.city/tags/%E5%8F%8C%E6%8C%87%E9%92%88/"/>
<category term="快慢指针" scheme="https://andy.city/tags/%E5%BF%AB%E6%85%A2%E6%8C%87%E9%92%88/"/>
<category term="字符串" scheme="https://andy.city/tags/%E5%AD%97%E7%AC%A6%E4%B8%B2/"/>
</entry>
<entry>
<title>LeetCode-151.反转字符串中的单词</title>
<link href="https://andy.city/archives/9090.html"/>
<id>https://andy.city/archives/9090.html</id>
<published>2022-10-04T12:47:04.000Z</published>
<updated>2022-10-04T12:47:04.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/reverse-words-in-a-string/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/0151.翻转字符串里的单词.html" target="_blank" rel="noopener">代码随想录</a></li></ul><iframe class="iframe_video" src="//player.bilibili.com/player.html?aid=471513250&bvid=BV1uT41177fX&cid=790645275&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen></iframe><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/0151.翻转字符串里的单词.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>要保持最优空间复杂度，只能原地修改字符串（js中字符串是不可变数据类型，只能先转为数组，后续操作的算法思想是正确的）</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><ul><li>移除多余空格时的边界条件处理</li></ul><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">reverseWords</span>(<span class="params">s: <span class="built_in">string</span></span>): <span class="title">string</span> </span>{</span><br><span class="line"> <span class="keyword">const</span> strArr = s.split(<span class="string">''</span>)</span><br><span class="line"> removeExtraSpaces(strArr)</span><br><span class="line"> <span class="keyword">const</span> len = strArr.length</span><br><span class="line"> <span class="comment">// 先整体反转</span></span><br><span class="line"> reverse(strArr, <span class="number">0</span>, len - <span class="number">1</span>)</span><br><span class="line"> <span class="comment">// 再把每个单词反转回来</span></span><br><span class="line"> <span class="keyword">let</span> start = <span class="number">0</span></span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">0</span>; i <= len; i++) {</span><br><span class="line"> <span class="keyword">if</span> (strArr[i] === <span class="string">' '</span> || i === len) {</span><br><span class="line"> reverse(strArr, start, i - <span class="number">1</span>)</span><br><span class="line"> start = i + <span class="number">1</span></span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> strArr.join(<span class="string">''</span>)</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="comment">/** 删除多余空格 */</span></span><br><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">removeExtraSpaces</span>(<span class="params">strArr: <span class="built_in">string</span>[]</span>) </span>{</span><br><span class="line"> <span class="keyword">let</span> slow = <span class="number">0</span>,</span><br><span class="line"> fast = <span class="number">0</span></span><br><span class="line"> <span class="keyword">while</span> (fast < strArr.length) {</span><br><span class="line"> <span class="comment">// 移除开始位置和重复的空格</span></span><br><span class="line"> <span class="keyword">if</span> (strArr[fast] === <span class="string">' '</span> && (fast === <span class="number">0</span> || strArr[fast - <span class="number">1</span>] === <span class="string">' '</span>)){</span><br><span class="line"> fast++</span><br><span class="line"> } <span class="keyword">else</span> {</span><br><span class="line"> strArr[slow++] = strArr[fast++]</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> strArr.length = strArr[slow - <span class="number">1</span>] === <span class="string">' '</span> ? slow - <span class="number">1</span> : slow</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="comment">/** 反转数组元素 */</span></span><br><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">reverse</span>(<span class="params">strArr: <span class="built_in">string</span>[], start: <span class="built_in">number</span>, end: <span class="built_in">number</span></span>) </span>{</span><br><span class="line"> <span class="keyword">while</span> (start < end) {</span><br><span class="line"> [strArr[start], strArr[end]] = [strArr[end], strArr[start]]</span><br><span class="line"> start++</span><br><span class="line"> end--</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n)<br>空间复杂度：O(n)</p><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="双指针" scheme="https://andy.city/tags/%E5%8F%8C%E6%8C%87%E9%92%88/"/>
<category term="快慢指针" scheme="https://andy.city/tags/%E5%BF%AB%E6%85%A2%E6%8C%87%E9%92%88/"/>
<category term="字符串" scheme="https://andy.city/tags/%E5%AD%97%E7%AC%A6%E4%B8%B2/"/>
</entry>
<entry>
<title>LeetCode-剑指Offer-05.替换空格</title>
<link href="https://andy.city/archives/56e6.html"/>
<id>https://andy.city/archives/56e6.html</id>
<published>2022-10-04T12:32:43.000Z</published>
<updated>2022-10-04T12:32:43.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/ti-huan-kong-ge-lcof/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/剑指Offer05.替换空格.html" target="_blank" rel="noopener">代码随想录</a></li></ul><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><ul><li>遍历一遍字符串，然后拼接</li></ul><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/剑指Offer05.替换空格.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>可以数组填充法（预先给数组扩容带填充后的大小，然后在从后向前进行操作），原地修改<ul><li>注意要从后向前填充，因为原数组会被膨胀（从前向后填充就是O(n^2)的算法了，因为每次添加元素都要将添加元素之后的所有元素向后移动）</li><li>类似于<a href="/archives/64f.html">27.移除元素</a>，我们是从前向后覆盖原数组，因为原数组是在做删除操作会被收缩</li></ul></li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><ul><li>注意要从后往前填充</li></ul><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">replaceSpace</span>(<span class="params">s: <span class="built_in">string</span></span>): <span class="title">string</span> </span>{</span><br><span class="line"> <span class="keyword">const</span> res = s.split(<span class="string">''</span>)</span><br><span class="line"> <span class="keyword">const</span> spaceNum = res.reduce(<span class="function">(<span class="params">acc, cur</span>) =></span> cur === <span class="string">' '</span> ? acc + <span class="number">1</span> : acc, <span class="number">0</span>)</span><br><span class="line"> <span class="keyword">let</span> left = s.length - <span class="number">1</span>,</span><br><span class="line"> right = left + spaceNum * <span class="number">2</span></span><br><span class="line"> <span class="keyword">while</span> (left >= <span class="number">0</span>) {</span><br><span class="line"> <span class="keyword">if</span> (res[left] === <span class="string">' '</span>) {</span><br><span class="line"> res[right--] = <span class="string">'0'</span></span><br><span class="line"> res[right--] = <span class="string">'2'</span></span><br><span class="line"> res[right--] = <span class="string">'%'</span></span><br><span class="line"> } <span class="keyword">else</span> {</span><br><span class="line"> res[right--] = res[left]</span><br><span class="line"> }</span><br><span class="line"> left--</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> res.join(<span class="string">''</span>)</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n)<br>空间复杂度：O(n)（这种解法的空间复杂度已经最优了。由于js中字符串是一个基本数据类型，即不可变类型，只能先转为数组再处理，后续操作的算法思想是正确的。如果字符串在你使用的编程语言中是一种可变数据类型，这种算法的空间复杂度就是O(1)）</p><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2><ul><li>将字符串转化为数组便于操作</li></ul>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="双指针" scheme="https://andy.city/tags/%E5%8F%8C%E6%8C%87%E9%92%88/"/>
<category term="字符串" scheme="https://andy.city/tags/%E5%AD%97%E7%AC%A6%E4%B8%B2/"/>
</entry>
<entry>
<title>LeetCode-541.反转字符串II</title>
<link href="https://andy.city/archives/c80d.html"/>
<id>https://andy.city/archives/c80d.html</id>
<published>2022-10-04T08:48:32.000Z</published>
<updated>2022-10-04T08:48:32.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/reverse-string-ii/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/0541.反转字符串II.html" target="_blank" rel="noopener">代码随想录</a></li></ul><iframe class="iframe_video" src="//player.bilibili.com/player.html?aid=471495028&bvid=BV1dT411j7NN&cid=787669511&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen></iframe><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><ul><li>模拟字符串遍历，分段拼接字符串</li></ul><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/0541.反转字符串II.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>可以将字符串先转化成数组，便于原地进行字符反转(同<a href="/archives/f3c.html">344.反转字符串</a>)，最后再转回字符串</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">reverseStr</span>(<span class="params">s: <span class="built_in">string</span>, k: <span class="built_in">number</span></span>): <span class="title">string</span> </span>{</span><br><span class="line"> <span class="keyword">const</span> res = s.split(<span class="string">''</span>)</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">0</span>, len = res.length; i < len; i += k * <span class="number">2</span>) {</span><br><span class="line"> <span class="keyword">let</span> left = i,</span><br><span class="line"> right = <span class="built_in">Math</span>.min(i + k - <span class="number">1</span>, len - <span class="number">1</span>)</span><br><span class="line"> <span class="keyword">while</span>(left < right) {</span><br><span class="line"> [res[left], res[right]] = [res[right], res[left]]</span><br><span class="line"> left++</span><br><span class="line"> right--</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> res.join(<span class="string">''</span>)</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n)<br>空间复杂度：O(n)</p><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2><ul><li>js中字符串是基本类型，不能直接修改，可转化为数组便于操作</li></ul>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="双指针" scheme="https://andy.city/tags/%E5%8F%8C%E6%8C%87%E9%92%88/"/>
<category term="字符串" scheme="https://andy.city/tags/%E5%AD%97%E7%AC%A6%E4%B8%B2/"/>
</entry>
<entry>
<title>LeetCode-344.反转字符串</title>
<link href="https://andy.city/archives/f3c.html"/>
<id>https://andy.city/archives/f3c.html</id>
<published>2022-10-04T08:29:00.000Z</published>
<updated>2022-10-04T08:29:00.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/reverse-string/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/0344.反转字符串.html" target="_blank" rel="noopener">代码随想录</a></li></ul><iframe class="iframe_video" src="//player.bilibili.com/player.html?aid=856322054&bvid=BV1fV4y17748&cid=785729810&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen></iframe><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><ul><li>双指针收缩，每次交换两个指针的元素值</li></ul><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/0344.反转字符串.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>思路一致</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">reverseString</span>(<span class="params">s: <span class="built_in">string</span>[]</span>): <span class="title">void</span> </span>{</span><br><span class="line"> <span class="keyword">let</span> left = <span class="number">0</span>,</span><br><span class="line"> right = s.length - <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> (left < right) {</span><br><span class="line"> [s[left], s[right]] = [s[right], s[left]]</span><br><span class="line"> left++</span><br><span class="line"> right--</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n)<br>空间复杂度：O(1)</p><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2><ul><li>在算法的解题思路上，字符串和数组是差不多的，只不过在不同编程语言中有着不同的操作</li></ul>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="双指针" scheme="https://andy.city/tags/%E5%8F%8C%E6%8C%87%E9%92%88/"/>
<category term="字符串" scheme="https://andy.city/tags/%E5%AD%97%E7%AC%A6%E4%B8%B2/"/>
</entry>
<entry>
<title>LeetCode-18.四数之和</title>
<link href="https://andy.city/archives/98c7.html"/>
<id>https://andy.city/archives/98c7.html</id>
<published>2022-10-04T07:52:12.000Z</published>
<updated>2022-10-04T07:52:12.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/4sum/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/0018.四数之和.html" target="_blank" rel="noopener">代码随想录</a></li></ul><iframe class="iframe_video" src="//player.bilibili.com/player.html?aid=728760817&bvid=BV1DS4y147US&cid=783669679&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen></iframe><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><ul><li>与<a href="/archives/4d32.html">15.三数之和</a>思路一样，外面多了一层循环</li></ul><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/0018.四数之和.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>彻底理解了双指针解法，与Carl思路一致</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><ul><li>去重</li><li>剪枝</li></ul><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">fourSum</span>(<span class="params">nums: <span class="built_in">number</span>[], target: <span class="built_in">number</span></span>): <span class="title">number</span>[][] </span>{</span><br><span class="line"> <span class="keyword">const</span> len = nums.length</span><br><span class="line"> <span class="keyword">if</span> (len < <span class="number">4</span>) <span class="keyword">return</span> []</span><br><span class="line"> nums.sort(<span class="function">(<span class="params">a, b</span>) =></span> a - b)</span><br><span class="line"> <span class="keyword">const</span> res: <span class="built_in">number</span>[][] = []</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">0</span>; i < len - <span class="number">3</span>; i++) {</span><br><span class="line"> <span class="keyword">if</span> (nums[i] > target && nums[i] > <span class="number">0</span> && target > <span class="number">0</span>) <span class="keyword">break</span> <span class="comment">// 剪枝</span></span><br><span class="line"> <span class="keyword">if</span> (i > <span class="number">0</span> && nums[i] === nums[i - <span class="number">1</span>]) <span class="keyword">continue</span> <span class="comment">// 去重</span></span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">let</span> j = i + <span class="number">1</span>; j < len - <span class="number">2</span>; j++) {</span><br><span class="line"> <span class="keyword">if</span> (nums[i] + nums[j] > target && nums[i] + nums[j] > <span class="number">0</span> && target > <span class="number">0</span>) <span class="keyword">break</span> <span class="comment">// 剪枝</span></span><br><span class="line"> <span class="keyword">if</span> (j > i + <span class="number">1</span> && nums[j] === nums[j - <span class="number">1</span>]) <span class="keyword">continue</span> <span class="comment">// 去重</span></span><br><span class="line"> <span class="keyword">let</span> left = j + <span class="number">1</span>,</span><br><span class="line"> right = len - <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> (left < right) {</span><br><span class="line"> <span class="keyword">const</span> sum = nums[i] + nums[j] + nums[left] + nums[right]</span><br><span class="line"> <span class="keyword">if</span> (sum < target) left++</span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span> (sum > target) right--</span><br><span class="line"> <span class="keyword">else</span> {</span><br><span class="line"> res.push([nums[i], nums[j], nums[left++], nums[right--]])</span><br><span class="line"> <span class="keyword">while</span> (left < right && nums[left] === nums[left - <span class="number">1</span>]) left++ <span class="comment">// 去重</span></span><br><span class="line"> <span class="keyword">while</span> (left < right && nums[right] === nums[right + <span class="number">1</span>]) right-- <span class="comment">// 去重</span></span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> res</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n^3)<br>空间复杂度：O(n)</p><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="双指针" scheme="https://andy.city/tags/%E5%8F%8C%E6%8C%87%E9%92%88/"/>
</entry>
<entry>
<title>LeetCode-15.三数之和</title>
<link href="https://andy.city/archives/4d32.html"/>
<id>https://andy.city/archives/4d32.html</id>
<published>2022-10-04T07:03:50.000Z</published>
<updated>2022-10-04T07:03:50.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/3sum/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/0015.三数之和.html" target="_blank" rel="noopener">代码随想录</a></li></ul><iframe class="iframe_video" src="//player.bilibili.com/player.html?aid=941244988&bvid=BV1GW4y127qo&cid=780823635&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen></iframe><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><p>之前做过的忘了。。。</p><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/0015.三数之和.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>巧用双指针，看完视频后对去重细节理解了，自己写就一遍过了</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><ul><li>对第一个数去重</li><li>对后两个数去重</li></ul><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">threeSum</span>(<span class="params">nums: <span class="built_in">number</span>[]</span>): <span class="title">number</span>[][] </span>{</span><br><span class="line"> <span class="keyword">const</span> len = nums.length;</span><br><span class="line"> <span class="keyword">if</span> (len < <span class="number">3</span>) <span class="keyword">return</span> []</span><br><span class="line"> <span class="keyword">const</span> res: <span class="built_in">number</span>[][] = []</span><br><span class="line"> nums.sort(<span class="function">(<span class="params">a, b</span>) =></span> a - b)</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">0</span>; i < len - <span class="number">2</span>; i++) {</span><br><span class="line"> <span class="keyword">if</span> (nums[i] > <span class="number">0</span>) <span class="keyword">break</span> <span class="comment">// 剪枝</span></span><br><span class="line"> <span class="keyword">if</span> (i > <span class="number">0</span> && nums[i] === nums[i - <span class="number">1</span>]) <span class="keyword">continue</span> <span class="comment">// 去重</span></span><br><span class="line"> <span class="keyword">let</span> left = i + <span class="number">1</span>,</span><br><span class="line"> right = len - <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> (left < right) {</span><br><span class="line"> <span class="keyword">const</span> sum = nums[i] + nums[left] + nums[right]</span><br><span class="line"> <span class="keyword">if</span> (sum < <span class="number">0</span>) left++</span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">if</span> (sum > <span class="number">0</span>) right--</span><br><span class="line"> <span class="keyword">else</span> {</span><br><span class="line"> res.push([nums[i], nums[left++], nums[right--]])</span><br><span class="line"> <span class="keyword">while</span> (left < right && nums[left] === nums[left - <span class="number">1</span>]) left++ <span class="comment">// 去重</span></span><br><span class="line"> <span class="keyword">while</span> (left < right && nums[right] === nums[right + <span class="number">1</span>]) right-- <span class="comment">// 去重</span></span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> res</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n^2)<br>空间复杂度：O(n)</p><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="双指针" scheme="https://andy.city/tags/%E5%8F%8C%E6%8C%87%E9%92%88/"/>
</entry>
<entry>
<title>LeetCode-383.赎金信</title>
<link href="https://andy.city/archives/9001.html"/>
<id>https://andy.city/archives/9001.html</id>
<published>2022-10-04T03:58:50.000Z</published>
<updated>2022-10-04T03:58:50.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/ransom-note/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/0383.赎金信.html" target="_blank" rel="noopener">代码随想录</a></li></ul><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><ul><li>哈希表记录每个字符串出现的次数</li></ul><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/0383.赎金信.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>思路一致</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><ul><li><p>Hash Array</p><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">canConstruct</span>(<span class="params">ransomNote: <span class="built_in">string</span>, magazine: <span class="built_in">string</span></span>): <span class="title">boolean</span> </span>{</span><br><span class="line"> <span class="keyword">const</span> hash = <span class="built_in">Array</span>(<span class="number">26</span>).fill(<span class="number">0</span>)</span><br><span class="line"> <span class="keyword">const</span> base = <span class="string">'a'</span>.codePointAt(<span class="number">0</span>)</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">0</span>; i < magazine.length; i++) hash[magazine.codePointAt(i) - base]++</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">0</span>; i < ransomNote.length; i++) {</span><br><span class="line"> <span class="keyword">const</span> index = ransomNote.codePointAt(i) - base</span><br><span class="line"> hash[index]--</span><br><span class="line"> <span class="keyword">if</span> (hash[index] < <span class="number">0</span>) <span class="keyword">return</span> <span class="literal">false</span></span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="literal">true</span></span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n)<br>空间复杂度：O(n)</p></li><li><p>Hash Map</p><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">canConstruct</span>(<span class="params">ransomNote: <span class="built_in">string</span>, magazine: <span class="built_in">string</span></span>): <span class="title">boolean</span> </span>{</span><br><span class="line"> <span class="keyword">const</span> hash = <span class="keyword">new</span> Map<<span class="built_in">string</span>, <span class="built_in">number</span>>()</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">const</span> c of magazine) hash.set(c, (hash.get(c) || <span class="number">0</span>) + <span class="number">1</span>)</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">const</span> c of ransomNote) {</span><br><span class="line"> <span class="keyword">const</span> count = hash.get(c)</span><br><span class="line"> <span class="keyword">if</span> (count > <span class="number">0</span>) hash.set(c, count - <span class="number">1</span>)</span><br><span class="line"> <span class="keyword">else</span> <span class="keyword">return</span> <span class="literal">false</span></span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="literal">true</span></span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n)<br>空间复杂度：O(n)</p></li></ul><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2><ul><li>已经轻车熟路了，与<a href="/archives/6aa1.html">242.有效的字母异位词</a>解法一致</li></ul>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="哈希表" scheme="https://andy.city/tags/%E5%93%88%E5%B8%8C%E8%A1%A8/"/>
</entry>
<entry>
<title>LeetCode-454.四数相加II</title>
<link href="https://andy.city/archives/4a6c.html"/>
<id>https://andy.city/archives/4a6c.html</id>
<published>2022-10-04T03:33:34.000Z</published>
<updated>2022-10-04T03:33:34.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/4sum-ii/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/0454.四数相加II.html" target="_blank" rel="noopener">代码随想录</a></li></ul><iframe class="iframe_video" src="//player.bilibili.com/player.html?aid=386006485&bvid=BV1Md4y1Q7Yh&cid=778751921&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen></iframe><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/0454.四数相加II.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>将数组分为两部分的思路很好，之前没想到</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">fourSumCount</span>(<span class="params">nums1: <span class="built_in">number</span>[], nums2: <span class="built_in">number</span>[], nums3: <span class="built_in">number</span>[], nums4: <span class="built_in">number</span>[]</span>): <span class="title">number</span> </span>{</span><br><span class="line"> <span class="keyword">let</span> res = <span class="number">0</span></span><br><span class="line"> <span class="keyword">const</span> hash = <span class="keyword">new</span> Map<<span class="built_in">number</span>, <span class="built_in">number</span>>()</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">const</span> v1 of nums1) {</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">const</span> v2 of nums2) {</span><br><span class="line"> <span class="keyword">const</span> v = v1 + v2</span><br><span class="line"> hash.set(v, (hash.get(v) || <span class="number">0</span>) + <span class="number">1</span>)</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">const</span> v1 of nums3) {</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">const</span> v2 of nums4) {</span><br><span class="line"> res += (hash.get(<span class="number">0</span> - (v1 + v2)) || <span class="number">0</span>)</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> res</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n^2)<br>空间复杂度：O(n)</p><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2><ul><li>哈希法的经典题目</li></ul>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="哈希表" scheme="https://andy.city/tags/%E5%93%88%E5%B8%8C%E8%A1%A8/"/>
</entry>
<entry>
<title>LeetCode-1.两数之和</title>
<link href="https://andy.city/archives/ce53.html"/>
<id>https://andy.city/archives/ce53.html</id>
<published>2022-10-04T02:45:34.000Z</published>
<updated>2022-10-04T02:45:34.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/two-sum/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/0001.两数之和.html" target="_blank" rel="noopener">代码随想录</a></li></ul><iframe class="iframe_video" src="//player.bilibili.com/player.html?aid=471096893&bvid=BV1aT41177mK&cid=776815010&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen></iframe><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><ul><li>我们需要记录每个元素出现的位置，所以用<code>hash map</code>即可</li></ul><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/0001.两数之和.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>思路一致</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><ul><li>要搞清楚哈希表的<code>key</code>、<code>value</code>分别是什么</li></ul><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">twoSum</span>(<span class="params">nums: <span class="built_in">number</span>[], target: <span class="built_in">number</span></span>): <span class="title">number</span>[] </span>{</span><br><span class="line"> <span class="keyword">const</span> hash = <span class="keyword">new</span> Map<<span class="built_in">number</span>, <span class="built_in">number</span>>()</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">0</span>; i < nums.length; i++) {</span><br><span class="line"> <span class="keyword">const</span> num = nums[i],</span><br><span class="line"> s = target - num</span><br><span class="line"> <span class="keyword">if</span> (hash.has(s)) <span class="keyword">return</span> [hash.get(s), i]</span><br><span class="line"> <span class="keyword">else</span> hash.set(num, i)</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> []</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n)<br>空间复杂度：O(n)</p><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2><ul><li><code>hash map</code>的经典应用</li></ul>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="哈希表" scheme="https://andy.city/tags/%E5%93%88%E5%B8%8C%E8%A1%A8/"/>
</entry>
<entry>
<title>LeetCode-202.快乐数</title>
<link href="https://andy.city/archives/971d.html"/>
<id>https://andy.city/archives/971d.html</id>
<published>2022-10-04T02:29:03.000Z</published>
<updated>2022-10-04T02:29:03.000Z</updated>
<content type="html"><![CDATA[<h2 id="题目详解"><a href="#题目详解" class="headerlink" title="题目详解"></a><a href="https://leetcode.cn/problems/happy-number/" target="_blank" rel="noopener">题目</a>详解</h2><h3 id="相关链接"><a href="#相关链接" class="headerlink" title="相关链接"></a>相关链接</h3><ul><li><a href="https://programmercarl.com/0202.快乐数.html" target="_blank" rel="noopener">代码随想录</a></li></ul><h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><ul><li>题目中说了如果不是快乐数就会出现无限循环，那么用哈希表记录n的每次迭代值：<ul><li>出现1则为快乐数</li><li>出现重复值则不是</li></ul></li></ul><h3 id="看完代码随想录之后的想法"><a href="#看完代码随想录之后的想法" class="headerlink" title="看完代码随想录之后的想法"></a>看完<a href="https://programmercarl.com/0202.快乐数.html" target="_blank" rel="noopener">代码随想录</a>之后的想法</h3><ul><li>思路一致</li></ul><h3 id="实现过程中遇到的困难"><a href="#实现过程中遇到的困难" class="headerlink" title="实现过程中遇到的困难"></a>实现过程中遇到的困难</h3><ul><li>计算迭代值时要细心点</li></ul><h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><ul><li><p>哈希表</p><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">isHappy</span>(<span class="params">n: <span class="built_in">number</span></span>): <span class="title">boolean</span> </span>{</span><br><span class="line"> <span class="keyword">const</span> hash = <span class="keyword">new</span> Set<<span class="built_in">number</span>>()</span><br><span class="line"> <span class="keyword">while</span> (n !== <span class="number">1</span>) {</span><br><span class="line"> <span class="keyword">if</span> (hash.has(n)) <span class="keyword">return</span> <span class="literal">false</span></span><br><span class="line"> hash.add(n)</span><br><span class="line"> n = getNext(n)</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="literal">true</span></span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">getNext</span>(<span class="params">n: <span class="built_in">number</span></span>): <span class="title">number</span> </span>{</span><br><span class="line"> <span class="keyword">let</span> sum = <span class="number">0</span></span><br><span class="line"> <span class="keyword">while</span> (n) {</span><br><span class="line"> sum += (n % <span class="number">10</span>) ** <span class="number">2</span></span><br><span class="line"> n = <span class="built_in">Math</span>.floor(n / <span class="number">10</span>)</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> sum</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n)<br>空间复杂度：O(n)</p></li><li><p>快慢指针（空间复杂度较好）</p><figure class="highlight ts"><figcaption><span>TypeScript</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">isHappy</span>(<span class="params">n: <span class="built_in">number</span></span>): <span class="title">boolean</span> </span>{</span><br><span class="line"> <span class="keyword">let</span> slow = n,</span><br><span class="line"> fast = n</span><br><span class="line"> <span class="comment">// 环形链表的思想，如果有环，快慢指针一定会在环内相遇</span></span><br><span class="line"> <span class="keyword">while</span> (<span class="literal">true</span>) {</span><br><span class="line"> slow = getNext(slow)</span><br><span class="line"> fast = getNext(getNext(fast))</span><br><span class="line"> <span class="keyword">if</span> (slow === fast) <span class="keyword">return</span> slow === <span class="number">1</span></span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">getNext</span>(<span class="params">n: <span class="built_in">number</span></span>): <span class="title">number</span> </span>{</span><br><span class="line"> <span class="keyword">let</span> sum = <span class="number">0</span></span><br><span class="line"> <span class="keyword">while</span> (n) {</span><br><span class="line"> sum += (n % <span class="number">10</span>) ** <span class="number">2</span></span><br><span class="line"> n = <span class="built_in">Math</span>.floor(n / <span class="number">10</span>)</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> sum</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>时间复杂度：O(n)<br>空间复杂度：O(1)</p></li></ul><h2 id="收获"><a href="#收获" class="headerlink" title="收获"></a>收获</h2><ul><li>要透过现象发现本质。本题看上去像是数学问题，其实不是。</li></ul>]]></content>
<summary type="html">
</summary>
<category term="算法" scheme="https://andy.city/categories/%E7%AE%97%E6%B3%95/"/>
<category term="LeetCode" scheme="https://andy.city/tags/LeetCode/"/>
<category term="代码随想录" scheme="https://andy.city/tags/%E4%BB%A3%E7%A0%81%E9%9A%8F%E6%83%B3%E5%BD%95/"/>
<category term="哈希表" scheme="https://andy.city/tags/%E5%93%88%E5%B8%8C%E8%A1%A8/"/>
<category term="双指针" scheme="https://andy.city/tags/%E5%8F%8C%E6%8C%87%E9%92%88/"/>
<category term="快慢指针" scheme="https://andy.city/tags/%E5%BF%AB%E6%85%A2%E6%8C%87%E9%92%88/"/>
</entry>
</feed>