First Bad Version Leetcode

最新推荐文章于 2025-02-03 15:37:10 发布

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本文介绍了一种使用二分查找算法来高效确定首次出现故障的产品版本的方法。通过最小化调用API次数来降低复杂度。

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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.

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

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解法：

最快找到，无非就是二分法；但这道题卡了好久，在某个比较大的测试用例同不过，正确代码如下：

public class Solution extends VersionControl {
    public int firstBadVersion(int n) {
        int start = 1,end = n;
        while(start < end){
            // int mid = (start + end) / 2; 
            // (start + end) >> 2
            // int mid = start + (end-start) / 2;??差别就在这个地方
            //'left + right' may cause the Integer Overflow, meaning that left+right > 2147483647
            int mid = start + (end-start) / 2;
            if(isBadVersion(mid)){
                end = mid;
            }
            else{
                start = mid + 1;
            }
        }
        return start;
    }
}

其中差别就在一句话：

int mid = (start + end) / 2;

和

int mid = start + (end -start) / 2;

导致异常问题已经查明：

'left + right' may cause the Integer Overflow, meaning that left+right > 2147483647

也就是说，加法在两个数比较大的时候会产生移除，因为两个数都是int类型