LeetCode 122. Best Time to Buy and Sell Stock II（greedy）

题目链接：https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/

题目描述：

Say you have an array for which the ith 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).

思路：这道题的思路需要好好思考一下对于每一笔交易 假设是从i天买入 j天卖出 ，如果有j+1天的价格高于j天，所以肯定是从i到j+1利润更高，同样的如果有第i-1天价格比第i天更低，肯定是从i-1天到j天利润更高。所以是不是对于每一段我们只用找到波谷和波峰，相减就可以了呢？
再分析一下，可以推出一个非常简单的式子，只要后项比前项大，这部分值就一定可以变为利润，即：
sum = sum+a[i+1]-a[i](if a[i+1]>a[i])，即贪心法

代码：

class Solution {public:    int maxProfit(vector<int>& prices) {        int n=prices.size();        int sum=0;        for(int i=1;i<n;i++)        {            if(prices[i]>prices[i-1])                sum+=prices[i]-prices[i-1];        }        return sum;    }};

贪心法介绍的链接：http://www.cnblogs.com/steven_oyj/archive/2010/05/22/1741375.html