博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
121. Best Time to Buy and Sell Stock(LeetCode)
阅读量:5343 次
发布时间:2019-06-15

本文共 968 字,大约阅读时间需要 3 分钟。

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:

Input: [7, 1, 5, 3, 6, 4]Output: 5max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

 

Example 2:

Input: [7, 6, 4, 3, 1]Output: 0In this case, no transaction is done, i.e. max profit = 0.
1 class Solution { 2 public: 3     int maxProfit(vector
& prices) { 4 int m = prices.size(); 5 if(m == 0) 6 return 0; 7 8 int curMin = prices[0]; 9 int ret = 0;10 for(int i = 1; i < m; i ++)11 {12 curMin = min(curMin, prices[i]);13 ret = max(ret, prices[i]-curMin);14 }15 return ret;16 }17 };

 

转载于:https://www.cnblogs.com/wujufengyun/p/6861399.html

你可能感兴趣的文章
Ajax : load()
查看>>
MySQL-EXPLAIN执行计划Extra解释
查看>>
Zookeeper概述
查看>>
Zookeeper一致性级别
查看>>
Linux远程登录
查看>>
Linux自己安装redis扩展
查看>>
HDU 1016 Prime Ring Problem(dfs)
查看>>
C#中结构体与字节流互相转换
查看>>
session和xsrf
查看>>
Linux目录结构
查看>>
luoguP3414 SAC#1 - 组合数
查看>>
五一 DAY 4
查看>>
(转)接口测试用例设计(详细干货)
查看>>
【译】SSH隧道:本地和远程端口转发
查看>>
图片点击轮播(三)-----2017-04-05
查看>>
直播技术细节3
查看>>
《分布式服务架构:原理、设计于实战》总结
查看>>
java中new一个对象和对象=null有什么区别
查看>>
字母和数字键的键码值(keyCode)
查看>>
IE8调用window.open导出EXCEL文件题目
查看>>