278. First Bad Version
题目:
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 version are also bad.
Suppose you have n versions [1, 2, ..., n] and you want to find out the first bad one, which causes all the following ones to be bad.
You are given an API bool isBadVersion(version) which will return whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.
链接: http://leetcode.com/problems/first-bad-version/
题解:
找到First Bad Version,这里我们使用Binary Search就可以了。
Time Complexity - O(logn), Space Complexity - O(1)
/* The isBadVersion API is defined in the parent class VersionControl.
boolean isBadVersion(int version); */ public class Solution extends VersionControl {
public int firstBadVersion(int n) {
int lo = 1, hi = n; while(lo <= hi) {
int mid = lo + (hi - lo) / 2;
if(isBadVersion(mid)) {
hi = mid - 1;
} else {
lo = mid + 1;
}
} return lo;
}
}
二刷:
二分搜索查找bad version的左边界。在找到badversion的时候我们让hi = mid - 1,否则lo = mid + 1, 最后返回lo就是第一个badversion的地方。
Java:
Time Complexity - O(logn), Space Complexity - O(1)
/* The isBadVersion API is defined in the parent class VersionControl.
boolean isBadVersion(int version); */ public class Solution extends VersionControl {
public int firstBadVersion(int n) {
if (n < 1) {
return 1;
}
int lo = 1, hi = n;
while (lo <= hi) {
int mid = lo + (hi - lo) / 2;
if (!isBadVersion(mid)) {
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return lo;
}
}
三刷:
Binary Search查找左边界。举个例子 {g, b, b, b, b}就很好理解了。lo和hi相等时hi--,所以我们最后返回lo就可以了
Java:
/* The isBadVersion API is defined in the parent class VersionControl.
boolean isBadVersion(int version); */ public class Solution extends VersionControl {
public int firstBadVersion(int n) {
if (n < 2) return 1;
int lo = 1, hi = n;
while (lo <= hi) {
int mid = lo + (hi - lo) / 2;
if (isBadVersion(mid)) hi = mid - 1;
else lo = mid + 1;
}
return lo;
}
}
Update:
/* The isBadVersion API is defined in the parent class VersionControl.
boolean isBadVersion(int version); */ public class Solution extends VersionControl {
public int firstBadVersion(int n) {
if (n < 1) return n;
int lo = 1, hi = n;
while (lo <= hi) {
int mid = lo + (hi - lo) / 2;
if (isBadVersion(mid)) hi = mid - 1;
else lo = mid + 1;
}
return lo;
}
}
Reference:
278. First Bad Version的更多相关文章
- leetcode 704. Binary Search 、35. Search Insert Position 、278. First Bad Version
704. Binary Search 1.使用start+1 < end,这样保证最后剩两个数 2.mid = start + (end - start)/2,这样避免接近max-int导致的溢 ...
- 【leetcode】278. First Bad Version
problem 278. First Bad Version solution1:遍历: // Forward declaration of isBadVersion API. bool isBadV ...
- 278. First Bad Version - LeetCode
Question 278. First Bad Version Solution 题目大意:产品有5个版本1,2,3,4,5其中下一个版本依赖上一个版本,即版本4是坏的,5也就是坏的,现在要求哪个版本 ...
- 【leetcode❤python】 278. First Bad Version
#-*- coding: UTF-8 -*-# The isBadVersion API is already defined for you.# @param version, an integer ...
- leetcode 278. First Bad Version
You are a product manager and currently leading a team to develop a new product. Unfortunately, the ...
- (medium)LeetCode 278.First Bad Version
You are a product manager and currently leading a team to develop a new product. Unfortunately, the ...
- Leetcode 278 First Bad Version 二分查找(二分下标)
题意:找到第一个出问题的版本 二分查找,注意 mid = l + (r - l + 1) / 2;因为整数会溢出 // Forward declaration of isBadVersion API. ...
- Java [Leetcode 278]First Bad Version
题目描述: You are a product manager and currently leading a team to develop a new product. Unfortunately ...
- 【easy】278. First Bad Version
有一系列产品,从某个开始其后都不合格,给定一个判断是否合格的函数,找出N个产品中第一个不合格的产品. 正确答案: // Forward declaration of isBadVersion API. ...
随机推荐
- 代码实现native2assci
public static void main(String[] args) { String unicode = ""; String s="用户名"; ch ...
- Windows Phone 8内存控制研究 之 LonglistSelector使用陷阱
最近工作中常常被问到如何降低WP内存使用,便再一次开始研究内存问题,首先发现了LonglistSelector使用的一个常见问题: 概述 若将Longlistselector 控件的ItemsSour ...
- UIScrollView缩放图片操作
要想ScrollView缩放,必须告诉缩放那个控件,它自身的大小是不会缩放的: 并且ScrollView只能缩放自己内部的子控件: 1:这时就要用到代理,代理告诉ScrollView缩放哪个控件.(设 ...
- Bootstrap入门三:页面排版
在Bootstrap中,页面的排版都是从全局的概念上出发,定制了主体文本.强调文本.标题.Code风格.按钮.表单.表格等格式,并运用CSS3的@font-face和伪元素一起实现了一套icon主题. ...
- hibernate tool连接oracle生成pojo和xml文件无法查询表解决办法
需要在hibernate的配置文件中增加 <property name="hibernate.default_schema">[username]</proper ...
- 新装Centos常见问题及解决方案
1.可以ping通,但无法通过ssh连接虚拟机的解决方案 虚拟机上装了一个 Linux 玩玩, 但在启动 Linux 后,在 Windows 中通过 Xshell 以 SSH 方式连接到 Linux ...
- 【Convert Sorted Array to Binary Search Tree】cpp
题目: Given an array where elements are sorted in ascending order, convert it to a height balanced BST ...
- gvim编辑文件到github乱码
with below _vimrc settings, code uploaded to GitHub will display with proper encoding set encoding=u ...
- 05.Hibernate多对多关联
前言:本文讲解使用Hibernate映射多对多关联关系,并使用多种方式映射多对多关联. 1.数据库表的多对多关系 本文根据学生信息表(tb_student)和教师信息表(tb_teac ...
- [转载]115个Java面试题和答案
不知道大家有没有这样的体会,就是找工作的时候不得不准备大量面试题,而工作的时间长了面试题里的精髓却忘的差不多了... 转载几篇Java面试的bolg,温故而知新,最重要的是常来看看. 1. http: ...