Leetcode Best Time to Buy and Sell Stock II

/** * 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;    }};