leetcode 278. First Bad Version（第一个坏的版本）

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本文介绍了一种使用二分搜索算法来高效确定首次出现瑕疵的产品版本的方法。在一个由好版本和坏版本组成的产品序列中，通过调用isBadVersion()函数，可以快速定位到导致后续版本皆为瑕疵的第一个版本。

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 returns whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.

Example 1:

Input: n = 5, bad = 4
Output: 4
Explanation:
call isBadVersion(3) -> false
call isBadVersion(5) -> true
call isBadVersion(4) -> true
Then 4 is the first bad version.

相当于一个数组0，0，…, 0, 1, …, 1，要找出第一个1的index，需要调用isBadVersion函数来看是0还是1。

思路：
二分搜索，left=1, right=n, 判断是1时mid = right - 1, 判断是0时left = mid + 1, 直到left > right, 返回left即可

    public int firstBadVersion(int n) {
        int left = 1;
        int right = n;
        while(left <= right) {
            int mid = left + (right - left) / 2;
            if(isBadVersion(mid)) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }