/**
* 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 ofthe 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).
*
* 题目要求能够多次买卖，可是同一时间仅仅能有一股在手里。也就是说在下一次买入的时候，必须卖出手中的股票。买入和卖出能够发生在vector的同一个
* 下标处。比方输入vector为[6, 9, 12, 8, 4, 11, 2。 1, 9] ，能够同一时候vector[i]这个点买入和卖出。
* 所以利润为(-6 + 9) + (-9 + 12) + (-4 + 11) + (-1 + 9)
* 这样就能够在每次上升子序列之前买入，在上升子序列结束的时候卖出。相当于能够获得全部的上升子序列的收益。
*/
class Solution {
public:
int maxProfit(vector<int> &prices) {
int i = 0;
int size = prices.size();
if(size < 2){
return 0;
}
int totalProfit = 0;
for(i = 1; i < size; i++){
if(prices[i] > prices[i - 1]){
totalProfit += prices[i] - prices[i - 1];
}
}
return totalProfit;
}
};
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