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
[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.
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.
Input: n = 1, bad = 1 Output: 1
1 <= bad <= n <= 231 - 1
There's some cool things about this problem. For instance, you may notice that:
mid = (start+end)) / 2;
causes overflows if you use this. That's because when start + end are around
INT_MAX it will cause an overflow since you can't add 2
Instead, use this method:
mid = start+(end-start)/2;
# The isBadVersion API is already defined for you. # @param version, an integer # @return a bool # def isBadVersion(version): class Solution(object): def firstBadVersion(self, n): """ :type n: int :rtype: int """ # this is a binary search, we need 2 pointers for this. left, right = 0, n - 1 while left <= right: mid = left + (right - left) / 2 if isBadVersion(mid) == False: left = mid + 1 else: right = mid - 1 return left
I suggest this article for studying binary search: