122. Best Time to Buy and Sell Stock II

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).

思路：

题目里面说可以做无数次交易，只要最后的profit最大就可以了。表面上看，无数次交易，似乎不容易。

但是，换个方式来想，只要今天的价格比前一天的价格高，那么就可以昨天买入然后今天卖出。如果明天的价格比今天高，那就今天卖出了以后再买入，然后再明天卖出。所以，就只需要比较当前的价格比前一天的价格，只要当前价格高于前一天，就可以把这个盈利算入最终的最大盈利里面去。

代码如下：

class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.empty()) return 0;

int maxProfit = 0;
for(int i = 0; i < prices.size() - 1; ++i){
if(prices[i + 1] > prices[i])
maxProfit += prices[i + 1] - prices[i];
}
return maxProfit;
}
};