LeetCode 121. Best Time to Buy and Sell Stock

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.

最基本的动态规划。

状态转移方程：

minValue保存已遍历的前I个元素的最小值

profit[i+1] = max(profit[i],prices[i+1]-minValues)

AC代码如下：

class Solution {public:    int maxProfit(vector<int>& prices)    {        if(prices.size() < 2)            return 0;        int profit = 0;        int minValue = prices[0];        for(int i = 1;i < prices.size();i++)        {            profit = max(profit,prices[i]-minValue);            minValue = min(minValue,prices[i]);        }        return profit;    }};