LeetCode First Bad Version278_c leetcode 278-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_38362049/article/details/81272060

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。在一个包含多个连续版本的产品中，一旦某个版本开始出现故障，其后续的所有版本也会受到影响。通过调用一个名为 isBadVersion 的 API 来判断版本是否存在问题，并利用二分查找法，可以快速定位到首个故障版本。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

Example:

Given n = 5, and version = 4 is the first bad version.

call isBadVersion(3) -> false
call isBadVersion(5) -> true
call isBadVersion(4) -> true

Then 4 is the first bad version.

思路

本体为二分查找的变种，适应于求最小花费的问题

查找第一个1

【0】【0】【0】【0】【1】【1】【1】【1】【1】【1】【max】

min mid

min:头指针

max:虚拟尾指针

mid = (min + max) / 2

调整:

如果arr[mid] == 1,min = mid + 1;

如果arr[max] != 1, max = mid;

如果min==max，找到结果

注意:这里的mid要更改为min+（max - min）/ 2，由于样例给的max值变大，那么（min+max）/2有可能会超界

// Forward declaration of isBadVersion API.
bool isBadVersion(int version);

int firstBadVersion(int n) {
    int min = 0, max = n, mid;
    while(min < max){
        mid = min + (max - min) / 2;
        if(isBadVersion(mid)) max = mid;
        else min = mid + 1;
    }
    return min;
}