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).

分析：贪心算法。如果可以交易多次，那么只要当前一个价格比前一个价格高我们就可以交易，最终可以获得最大利益。代码如下：

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

Leetcode:Best Time to Buy and Sell Stock II