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).

题目：

最佳时间买入卖出股票：你有一个数组保存了股票在第i天的价钱，现在你可以进行多次买入卖出，但同一时间你手上只能保持一个股票，如何赚的最多

思路：

贪心法，本题和前面的Best Time to Buy and Sell Stock 不同在于，本题可以多次买卖股票，
从而累积赚取所有的价格差。因此用贪心法，基本思想是锁定一个低价，然后在价格升到局部最高点
（即下一天的价钱就下降了）时候，抛出股票，然后把下一天较低的价钱作为买入，接着计算。
要注意最后要处理最后一次的利润。

C++代码实现：

#include<iostream>#include<vector>using namespace std;class Solution{public:    int maxProfit(vector<int> &prices)    {        if(prices.empty())            return 0;        int buy=prices[0];        int sum=0;        int i;        for(i=1;i<(int)prices.size();i++)        {            if(buy<prices[i])            {                sum+=prices[i]-buy;                buy=prices[i];            }            else                buy=prices[i];        }            return sum;    }};int main(){    Solution s;    vector<int> vec={4,7,5,9};    cout<<s.maxProfit(vec)<<endl;}

Best Time to Buy and Sell Stock II