278. First Bad Version

问题描述

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 APIbool 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.

题目链接：

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思路分析

找到第一个bad version。

很tricky的一点，为了找到第一个bad version，我们实际要找的应该是最后一个good version，否则会无法判断的。因此有一个小改动，max不再-1，而是直接等于mid，这样min > max是就正好是第一个bad version了。

其次，不用(min + max)/2 的原因是可能会越界，因此采用的是min + (max - min)/2的方式。

代码

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

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

时间复杂度：$O(logn)$
空间复杂度：$O(1)$

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反思

实际上题目不难，二分查找，但是要注意小细节，考察对查找的理解。