题目链接
tag:

• Easy；

question：
  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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note: 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.

思路：
  比较简单，只需要遍历一次数组，用一个变量记录遍历过程中的买入股票最小值，然后每次计算当前值和这个最小值之间的差值为利润，然后每次选较大的利润来更新。当遍历完成即为最大利润，代码如下：

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;
    }
};