题目地址：

https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/

题目描述：

...

解决方案：

当前值减之前的数中的最小值得当前值最大利润，再更新最大利润即可

代码：

class Solution {public:    int maxProfit(vector
     
      & prices) {        if (prices.size() < 2) {            return 0;        }        int minPrice = prices.at(0);        int maxProfit = 0;        for (size_t i = 1; i < prices.size(); i++)        {            if (minPrice > prices.at(i)) {                minPrice = prices.at(i);                continue;            }            int profit = prices.at(i) - minPrice;            if (profit > 0 && profit > maxProfit) {                maxProfit = profit;            }                }        return maxProfit;    }};