Best Time to Buy and Sell Stock

一、问题描述

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.

二、思路

动态规划法求解。

在双层循环中，记录当前出现过的最低价格min_price，作为买入价格，并计算以当天价格出售的收益，作为可能的最大收益max_price，整个遍历过程中，出现过的最大收益max_price就是所求。

三、代码

class Solution {public:    int maxProfit(vector<int> &prices) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (prices.size() == 0)            return 0;                    int maxPrice = prices[prices.size()-1];        int ans = 0;        for(int i = prices.size() - 1; i >= 0; i--)        {            maxPrice = max(maxPrice, prices[i]);            ans = max(ans, maxPrice - prices[i]);        }                return ans;    }};