LeetCode 122 Best Time to Buy and Sell Stock II

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本文介绍了一种算法，用于解决无限次买卖股票以获取最大利润的问题。通过使用贪心算法，文章详细阐述了如何根据每日股价变化来确定最佳买卖时机。

题目

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).

无限次买卖股票，求赚取最大值

思路

1 马上想到贪心算法。

2 最终求的最大值其实是由每部分最大值全部加起来而成；所以只要碰到下一格上涨了，马上加上去，并且更新最低值；如果下跌了，马上更新最低值。

3 可以和http://blog.youkuaiyun.com/xift810/article/details/21647545 进行对比。

代码

public class Solution {
    public int maxProfit(int[] prices) {
        if(prices.length<=1){
            return 0;
        }
        
        int maxprofit = 0;
        int buy = prices[0];
        for(int i=1;i<prices.length;i++){
            if(prices[i]>buy){
                maxprofit += prices[i]-buy;
                buy = prices[i];
            }
            else{
                buy = prices[i];
            }
        }
        
        return maxprofit;
    }
}