从零开始的LC刷题(66): First Bad Version

最新推荐文章于 2024-07-15 09:53:28 发布

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最新推荐文章于 2024-07-15 09:53:28 发布

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分类专栏： LEETCODE C++ 文章标签： lc

本文链接：https://blog.youkuaiyun.com/cyr429/article/details/92650764

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博客围绕查找首个不良版本问题展开，已知产品各版本基于前一版本开发，有一个不良版本会导致后续版本都不良。给出一个判断版本是否不良的API，需实现函数找出首个不良版本并最小化API调用次数，还展示了C++代码运行结果。

原题：

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. 

依题意能感觉要用二分查找，但是这道题应该先试探再二分，另外二分也不能直接(a+b)/2因为中间有testcase会超出int范围，结果：

Success

Runtime: 0 ms, faster than 100.00% of C++ online submissions for First Bad Version.

Memory Usage: 8.1 MB, less than 65.27% of C++ online submissions for First Bad Version.

代码：

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

class Solution {
public:
    int firstBadVersion(int n) {
        int l=0,r=n;
        while(true){
            if(r-l==1){return r;}
            int temp=(r-l)/2+l;
            if (isBadVersion(temp)){r=temp;}
            else {l=temp;}
        }
        return 0;
    }
};