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.

题目大意：

假设你有一个数组，其中第i 个元素是第i天给定股票的价格。
如果只允许最多完成一个交易（即购买一个交易并且卖出一个股票），则设计一个算法来找到最大利润。

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

解题思路

遍历一边数组即可，分别记录下面的两个信息
1、用一个变量记录遍历过数中最小的一个数
2、每次计算当前值和这个最小值之间的差值作为利润，每次选择利润较大的数即可

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int res = 0, buy = INT_MAX;
        for (int price:prices)
        {
            buy = min(buy, price);
            res = max(res, price-buy);
        }
        
        return res;
    }
};
```