278. First Bad Version

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. 

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从[1,2,…,n]有n个版本，其中有个版本是坏的，导致其后的所有版本都是坏的，现在要找到第一个坏了的版本。
简单题，二分搜索，相当于找lowerbound，正好刷了LeetCode Search for a Range，理清了二分搜索查找上下界的方法，此题直接实现lowerbound就好。代码如下：

// Forward declaration of isBadVersion API. bool isBadVersion(int version);
class Solution {
public:
    int firstBadVersion(int n)
    {
        int l = 1, r = n;
        while (l <= r) {
            int m = l + (r – l) / 2;
            if (!isBadVersion(m))
                l = m + 1;
            else
                r = m – 1;
        }
        return l;
    }
};

本代码提交AC，用时0MS。

二刷：

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

本代码提交AC，用时0MS。