/**
 * 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).
 * 
 * 题目要求可以多次买卖，但是同一时间只能有一股在手里。也就是说在下一次买入的时候，必须卖出手中的股票。买入和卖出可以发生在vector的同一个
 * 下标处，比如输入vector为[6, 9, 12, 8, 4, 11, 2， 1, 9] ，可以同时vector[i]这个点买入和卖出。
 * 所以利润为(-6 + 9) + (-9 + 12) + (-4 + 11) + (-1 + 9)
 * 这样就可以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。相当于能够获得所有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;
    }
};

Leetcode Best Time to Buy and Sell Stock II