LeetCode 之 Best Time to Buy and Sell Stock II

原题

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).

这个题和1有点区别，可以看看这个 Best Time to Buy and Sell Stock ，本题还是很简单的，其实就是考虑两个相邻的数值，要是后一个比前一个大就可以算做一次交易，即在上一次低的时候买入，本次高的时候卖出，可以看个图。。。

源代码（44ms）

class Solution {public:    int maxProfit(vector<int> &prices) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        int total = 0;        if (prices.size() == 0) return 0;        int last = prices[0];                for(int i =1;i<prices.size();i++){           //只要比上一次挣钱做一次交易 。。。这里相当于买了last卖了prices[i]            if(prices[i]>prices[i-1]){                total += prices[i]-prices[i-1];            }        }                return total;    }};