121. Best Time to Buy and Sell Stock 类别：动态规划 难度：easy

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.

思路：

记ret为收益，随着prices的遍历，如果当前遇到的元素比curMin小，就重新更新curMin，计算prices[i]与curMin的差值，并更新ret。

程序：

class Solution {public:    int maxProfit(vector<int>& prices) {                int m = prices.size();        if(m == 0)            return 0;                    int curMin = prices[0];        int ret = 0;        for(int i = 1; i < m; i ++)        {            curMin = min(curMin, prices[i]);            ret = max(ret, prices[i]-curMin);        }        return ret;                    }};