121. Best Time to Buy and Sell Stock.md

121. Best Time to Buy and Sell Stock

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leetcode Daily Challenge on September 18th, 2020.

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Difficulty : Easy

Related Topics : Array、Dynamic Programming

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Say you have an array for which the i^th element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
             Not 7-1 = 6, as selling price needs to be larger than buying price.

Example 2:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

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Solution

• mine

  • Java

    Runtime: 1 ms, faster than 99.12%, Memory Usage: 39.7 MB, less than 29.47% of Java online submissions

    // O(N)time
    // O(1)space
    public int maxProfit(int[] prices) {
        if(prices == null || prices.length <= 1){
            return 0;
        }
        int res = 0;
        int min = Integer.MAX_VALUE;
        for(int i = 0; i < prices.length; i++){
            min = Math.min(min, prices[i]);
            res = Math.max(res, prices[i] - min);
        }
        return res;
    }
    

    Runtime: 241 ms, faster than 14.04%, Memory Usage: 39.4 MB, less than 63.87% of Java online submissions

    // O(N^2)time
    // O(1)space
    public int maxProfit(int[] prices) {
        if(prices == null || prices.length == 0){
            return 0;
        }
        int max = 0;
        int len = prices.length;
        for(int i = 0; i < len; i++){
            for(int j = i + 1; j < len; j++){
                max = Math.max(prices[j]  - prices[i], max);
            }    
        }
        return max;
    }
    

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• the most votes

• Runtime: 1 ms, faster than 99.12%, Memory Usage: 39.5 MB, less than 42.77% of Java online submissions

  // O(N)time
  // O(1)space
  public int maxProfit(int prices[]) {
      int minprice = Integer.MAX_VALUE;
      int maxprofit = 0;
      for (int i = 0; i < prices.length; i++) {
          if (prices[i] < minprice)
              minprice = prices[i];
          else if (prices[i] - minprice > maxprofit)
              maxprofit = prices[i] - minprice;
      }
      return maxprofit;
  }
  

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