best-time-to-buy-and-sell-stock-ii

题目：

Say you have an array for which the i th element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

程序：

class Solution {public:    int maxProfit(vector<int> &prices) {        //本题由于允许多次交易（每次必须先卖出再买进），所以不好用爆搜        //分析可知，要想利益最大化，就应该每次在波谷买进，波峰卖出，这样利益最大，操作次数最少        //应该是使用动态规划来做可能比较简洁，个人觉得。                 int len = prices.size();        vector<int> change(len,0);        int maxPro=0;        for(int i=1;i<len;i++){            change[i]=prices[i]-prices[i-1];  //记录所有长和跌的情况            if(change[i]>0)maxPro += change[i];  //累加所有长幅，即为最大收益        }        return maxPro;    }};