You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad ve…

Compare two version numbers version1 and version1.If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0. You may assume that the version strings are non-empty and contain only digits and the . character.The . characte…

题目描述: Compare two version numbers version1 and version2. If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0. You may assume that the version strings are non-empty and contain only digits and the . character. The .…

You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad ve…

Compare two version numbers version1 and version2.If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0. You may assume that the version strings are non-empty and contain only digits and the . character.The . characte…

Compare two version numbers version1 and version2.If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0. You may assume that the version strings are non-empty and contain only digits and the . character.The . characte…

Compare two version numbers version1 and version2.If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0. You may assume that the version strings are non-empty and contain only digits and the . character.The . characte…

题意: 有一个bool序列表示对应下标的版本是否出问题(下标从1开始),如果一个版本出了问题,那么其后面全部版本必定出问题.现在给出判断任意版本是否出问题的API,请找到第一个出问题的版本. 思路: 明显的二分查找. // Forward declaration of isBadVersion API. bool isBadVersion(int version); class Solution { public: int firstBadVersion(int n) { , R=n; whil…

Compare two version numbers version1 and version2.If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0. You may assume that the version strings are non-empty and contain only digits and the . character.The . characte…

题目描述: Compare two version numbers version1 and version2.If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0. You may assume that the version strings are non-empty and contain only digits and the . character.The . ch…

Total Accepted: 12400 Total Submissions: 83230     Compare two version numbers version1 and version2. If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0. You may assume that the version strings are non-empty and co…

# The isBadVersion API is already defined for you. # @param version, an integer # @return a bool # def isBadVersion(version): class Solution(object): def firstBadVersion(self, n): """ :type n: int :rtype: int """ if n < 1:…

You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad ve…

Description: You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions a…

问题描述: You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a…

题意:比较版本号的大小 有点变态,容易犯错 本质是字符串的比较,请注意他的版本号的小数点不知1个,有的会出现01.0.01这样的变态版本号 class Solution { public: int cmp_num(string a, string b){ ; else if(a.size() == b.size()){ ; : ; } ; } vector<string> change(string v){ vector<string> vs; , b = ; while((b =…

You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad ve…

大家好,我是 程序员小熊 ,来自 大厂 的程序猿.相信绝大部分程序猿都有一个进大厂的梦想,但相较于以前,目前大厂的面试,只要是研发相关岗位,算法题基本少不了,所以现在很多人都会去刷 Leetcode 来保持手感,但有不少人反馈刷题效率很低,今天笔者抽空整理了 三份 分别来自 谷歌的高畅.前阿里的霜神和灵魂机器 的刷题手册,以供大家参考,希望对大家无有所帮助. 一. A LeetCode Grinding Guide (C++ Version) 作者:谷歌的高畅 背景:作者在美国卡内基梅隆大学攻读…

Profile Introduction to Blog 您能看到这篇博客导读是我的荣幸.本博客会持续更新.感谢您的支持.欢迎您的关注与留言.博客有多个专栏,各自是关于 Android应用开发 .Windows App开发 . UWP(通用Windows平台)开发 . SICP习题解 和 Scheme语言学习 . 算法解析 与 LeetCode等题解 .而近期会加入的文章将主要是算法和Android.只是其他内容也会继续完好. About the Author 独立 Windows App 和…

Compare two version numbers version1 and version1.If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0. You may assume that the version strings are non-empty and contain only digits and the . character.The . characte…

原题链接在这里:https://leetcode.com/problems/compare-version-numbers/ 用string.split()方法把原有string 从小数点拆成 string 数组,但这里要注意 . 和 * 是不能直接用split(".") 或者split("*")拆开的,因为 . 可以代表任意char, * 可以代表任意字符串.所以要加 \\. 来避免individual special character. 拆开后用Interge…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 来源: https://leetcode.com/problems/compare-version-numbers/ Compare two version numbers version1 and version2. If version1 > version2 return 1, if version1 <…

这是悦乐书的第200次更新,第210篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第66题(顺位题号是278).您是产品经理,目前领导团队开发新产品.不幸的是,您产品的最新版本未通过质量检查.由于每个版本都是基于以前的版本开发的,因此坏版本之后的所有版本也是坏的. 假设您有n个版本[1,2,...,n]并且您想找出第一个坏的版本,这会导致以下所有版本都不好.您将获得一个API bool isBadVersion(版本),它将返回版本是否错误. 实现一个函数来查找第一…

[leetcode 字符串处理]Compare Version Numbers @author:wepon @blog:http://blog.csdn.net/u012162613 1.题目 Compare two version numbers version1 and version1. If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0. You may assume…

Leetcode之二分法专题-278. 第一个错误的版本(First Bad Version) 你是产品经理,目前正在带领一个团队开发新的产品.不幸的是,你的产品的最新版本没有通过质量检测.由于每个版本都是基于之前的版本开发的,所以错误的版本之后的所有版本都是错的. 假设你有 n 个版本 [1, 2, ..., n],你想找出导致之后所有版本出错的第一个错误的版本. 你可以通过调用 bool isBadVersion(version) 接口来判断版本号 version 是否在单元测试中出错.实现…

278. 第一个错误的版本 LeetCode278. First Bad Version 题目描述 你是产品经理,目前正在带领一个团队开发新的产品.不幸的是,你的产品的最新版本没有通过质量检测.由于每个版本都是基于之前的版本开发的,所以错误的版本之后的所有版本都是错的. 假设你有 n 个版本 [1, 2, ..., n],你想找出导致之后所有版本出错的第一个错误的版本. 你可以通过调用 bool isBadVersion(version) 接口来判断版本号 version 是否在单元测试中出错.…

问题 该文章的最新版本已迁移至个人博客[比特飞],单击链接 https://www.byteflying.com/archives/3985 访问. 你是产品经理,目前正在带领一个团队开发新的产品.不幸的是,你的产品的最新版本没有通过质量检测.由于每个版本都是基于之前的版本开发的,所以错误的版本之后的所有版本都是错的. 假设你有 n 个版本 [1, 2, ..., n],你想找出导致之后所有版本出错的第一个错误的版本. 你可以通过调用 bool isBadVersion(version) 接口来…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 解题方法 二分查找 日期 题目地址:https://leetcode.com/problems/first-bad-version/description/ 题目描述 You are a product manager and currently leading a team to develop a new product. Unfortunately, the la…

[LeetCode]165. Compare Version Numbers 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.me/ 题目地址:https://leetcode.com/problems/compare-version-numbers/description/ 题目描述: Compare two version numbers version1 and vers…

#-*- coding: UTF-8 -*-# The isBadVersion API is already defined for you.# @param version, an integer# @return a bool# def isBadVersion(version):class Solution(object):    def firstBadVersion(self, n):        """        :type n: int        :…