278. First Bad Version (二分，java)

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本文介绍了一种高效查找软件中首次出现故障版本的方法，利用二分查找算法减少API调用次数，确保快速定位问题。

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;
    }
}