LeetCode题解:Best Time to Buy and Sell Stock II
2015-10-02 21:07
337 查看
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).
题意:买卖股票的最佳时机,但这次能够完成任意数量的交易,但不能同时进行多项交易,即在买之前必须先卖。
解决思路:
代码:能赚就买卖……
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).
题意:买卖股票的最佳时机,但这次能够完成任意数量的交易,但不能同时进行多项交易,即在买之前必须先卖。
解决思路:
代码:能赚就买卖……
public class Solution { public int maxProfit(int[] prices) { int max = 0; for(int i = 0; i < prices.length - 1; i++){ if(prices[i + 1] > prices[i]){ max += prices[i + 1] - prices[i]; } } return max; } }
相关文章推荐
- Swift学习(二)
- 小坑+1 runAction 顺序
- C#学习之路 , 学习笔记 2.4 转义字符 和 @控制符
- Codeforces Round #274 (Div. 2)
- repo 获取Android源码
- LeetCode题解:Best Time to Buy and Sell Stock
- TCP/IP详解卷1 读书笔记:第七章 Ping程序
- Java学习笔记----封装和private关键字
- Android五个布局
- http://jingyan.baidu.com/article/a378c960630e61b329283045.html
- [codevs2152]滑雪
- 新文章
- LeetCode题解:Triangle
- [LeetCode]Longest Consecutive Sequence
- RPC漏洞
- CentOS 7更改网卡名称
- equal和==的区别(原理讲解)
- 大数乘法(快速傅立叶变换)下
- 三色旗
- Mysql命令大全