LeetCode122: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 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).

Hide Tags Array Greedy

这可以说是股票交易中比较简单的一种情况了。
从一个时间点开始算起，只要接下来一天的价格大于当天的价格，就进行一笔买卖。这样求出这些天所有的差价和就是最大的收益。
看了Hide Tags的提示发现这其实是一种贪婪思想，因为正好局部最优解就是全局最优解，所以可以用贪婪算法解题。
runtime:10ms

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