[LeetCode] Best Time to Buy and Sell Stock II Solution

本文介绍了一种解决无限次股票交易问题的算法,通过扫描股价数组并计算所有正价差之和来确定最大利润。该方法允许在进行多次买卖操作的同时避免同时持有股票。

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).

» Solve this problem

[Thoughts]
A bit different with previous one. Since we can make unlimited transactions, this question turns to sum all the positive price difference.

So, scan from left to right, and add all positive diff value.

[Code]

1:    int maxProfit(vector<int> &prices) {  
2: // Start typing your C/C++ solution below
3: // DO NOT write int main() function
4: int max=0;
5: int sum = 0;
6: for(int i =1; i< prices.size(); i++)
7: {
8: int diff = prices[i] -prices[i-1];
9: if(diff>0)
10: sum+=diff;
11: }
12: return sum;
13: }
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值