278. First Bad Version (二分，java)

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原创最新推荐文章于 2026-06-17 09:39:28 发布 493 阅读 AI 写同款· GEO 优化 ›

2022年电赛F题23年电赛D题-信号调制度测量装置说明中提到带通采样定律。另外，当对于一个频率很高的射频信号采样时，如果采样频率设的太低，对提高采样量化的信噪比是不利的。低通采样定理在《信号与系统》教材中进行了详细的讨论，但未对带通信号的采样定理进行描述，国内樊昌信和曹丽娜老师编写的《通信原理》第10章第2小节有所描述，但没有详细的公式推导，容易给初学者产生超过该最低采样频率即不会产生频谱混叠的假象。故采样频率没选好，就会使周期性频谱搬移过程中造成频谱混叠，既然频谱和原信号的频谱对不上了，故就无法无失真恢复原信号了，总之，频域波形不出现重叠是无失真传输的重要保障。阅读详情

278. First Bad Version

Description Submissions Solutions

Total Accepted: 90906
Total Submissions: 367256
Difficulty: Easy
Contributor: LeetCode

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.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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题目链接：https://leetcode.com/problems/first-bad-version/#/description

解题思路，这题对时间要求很高，原来我写二分时，mid都是写成：mid = ( high + low) / 2 ，但是一直超时，还有一种写法是 mid = low + ( high － low ) / 2 ，虽然计算结果是一样的，但后一种写法更好，原因在于：

1、第一种方法可能会出现溢出的风险，第二种写法不会出现比high大的数字，所以比较安全。

2、第二种的通用性更强，指针和迭代器运算不支持相加运算，却支持相减运算。

/* 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 lw = 1,hg = n;
        while(lw<hg){
            int mid =lw + (hg-lw)/2;
            if(isBadVersion(mid)){
                hg = mid;
            }else{
                lw = mid+1;
            }
        }
        return lw;
    }
}