LeetCode - Easy - 278

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 returns whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.

Example 1:

Input: n = 5, bad = 4

Output: 4

Explanation:

call isBadVersion(3) -> false

call isBadVersion(5) -> true

call isBadVersion(4) -> true

Then 4 is the first bad version.

Example 2:

Input: n = 1, bad = 1

Output: 1

Constraints:

  • 1 <= bad <= n <= 2³¹ - 1

Analysis


Submission


class VersionControl{

public static int badVersion;

public boolean isBadVersion(int version) {

return version >= badVersion;

}

}

public class FirstBadVersion extends VersionControl{

public int firstBadVersion(int n) {

int left = 1, right = n;

while(left < right) {

int mid = left + (right - left) / 2;

if(isBadVersion(mid)) {

right = mid;

}else {

left = mid + 1;

}

}

return right;

}

}

Test


import static org.junit.Assert.*;

import org.junit.Test;

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