121. Best Time to Buy and Sell Stock

题目：

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

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

Example 1:

Input: [7, 1, 5, 3, 6, 4]Output: 5max. difference = 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: 0In this case, no transaction is done, i.e. max profit = 0.

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思路：从数组的开始到结尾遍历数组，记录从开始到数组A[i]位最小的元素，这样后面比A[i]大的元素k只需与A[i]做差与之前的最大利润比较就可以了，不需要管在A[i]之前且比A[i]大的元素，因为一定比k-A[i]大。

代码：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int result = 0;     //存放遍历过程中的最大利润
        int temp = INT_MAX;        //存放遍历过程中的最小值
        for(int i = 0;i<prices.size();i++)
        {
            temp = min(temp, prices[i]);        //比较prices[i]和temp那个小。
            result = max(result, prices[i]-temp);         //只要利润大于0，就和之前的最大利润进行比较。
        }
        return result;
    }
};