给定一个字符串数组 words
,请计算当两个字符串 words[i]
和 words[j]
不包含相同字符时,它们长度的乘积的最大值。假设字符串中只包含英语的小写字母。如果没有不包含相同字符的一对字符串,返回 0。
示例 1:
输入: words =["abcw","baz","foo","bar","fxyz","abcdef"]
输出:16 解释: 这两个单词为
"abcw", "fxyz"
。它们不包含相同字符,且长度的乘积最大。
示例 2:
输入: words =["a","ab","abc","d","cd","bcd","abcd"]
输出:4 解释:
这两个单词为"ab", "cd"
。
示例 3:
输入: words =["a","aa","aaa","aaaa"]
输出:0 解释: 不存在这样的两个单词。
提示:
2 <= words.length <= 1000
1 <= words[i].length <= 1000
words[i]
仅包含小写字母
注意:本题与主站 318 题相同:https://leetcode-cn.com/problems/maximum-product-of-word-lengths/
因为只有 26 个小写字符,所以可以用一个 int32
存储字符的出现情况,然后枚举最大乘积
class Solution:
def maxProduct(self, words: List[str]) -> int:
n = len(words)
mask = [0 for _ in range(n)]
for i, word in enumerate(words):
for ch in word:
mask[i] |= 1 << (ord(ch) - ord('a'))
ans = 0
for i in range(0, n - 1):
for j in range(i + 1, n):
if mask[i] & mask[j] == 0:
ans = max(ans, len(words[i]) * len(words[j]))
return ans
class Solution {
public int maxProduct(String[] words) {
int n = words.length;
int[] mask = new int[n];
for (int i = 0; i < n; i++) {
for (char ch : words[i].toCharArray()) {
mask[i] |= 1 << (ch - 'a');
}
}
int ans = 0;
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
if ((mask[i] & mask[j]) == 0) {
ans = Math.max(ans, words[i].length() * words[j].length());
}
}
}
return ans;
}
}
func maxProduct(words []string) int {
n := len(words)
mask := make([]int32, n)
for i, word := range words {
for _, r := range word {
mask[i] |= 1 << (r - 'a')
}
}
ans := 0
for i := 0; i < n-1; i++ {
for j := i + 1; j < n; j++ {
if mask[i]&mask[j] == 0 {
ans = max(ans, len(words[i])*len(words[j]))
}
}
}
return ans
}
func max(a, b int) int {
if a > b {
return a
}
return b
}