Best Time to Buy and Sell Stock II问题及解法

问题描述：

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).

问题分析：
本题与之前的Best Time to Buy and Sell Stock的问题类似，但是这里又多了一个条件，就是股票可以买卖多次。

在这里，要想赚取更多的利益，每次卖出前，都要保证售出价>收购价。因此，我们只需要计算相邻的两个元素之间之差就可以。

过程详见代码：

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