【leetcode】122. Best Time to Buy and Sell Stock II

Difficulty：Medium

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 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).

解题思路（时间复杂度O(n)）：

对于题目中要求的可以进行无数次交易，那么其实使用贪心算法就可以很好地解决，对给出的数组进行一次遍历，如果prices[i]>prices[i-1]，那么说明在这两个值的差额是可以获得收益的，直到出现prices[i]<prices[i-1]那么之前的每一次差额都可以计算进总收益之内的。而在这个时候，相当于更新了min的值，再次进行之前的步骤，直到最后累加得到最大的收益

代码如下：

class Solution {public:    int maxProfit(vector<int>& prices) {        int ans=0;        int n=prices.size();        if(n<2) return 0;        for(int i=0;i<n-1;i++)        {            if(prices[i]<prices[i+1]) ans+=prices[i+1]-prices[i];        }        return ans;    }};