[LeetCode] 278. First Bad Version_Easy tag: Binary Search

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. 

这题思路实际上就跟[LeetCode] 374. Guess Number Higher or Lower_Easy tag: Binary Search很像, 就是找第一个bad version. 

04/14/2019 update: 或者说是找 first index isBadVersion（）结果为true。

 

Code

# The isBadVersion API is already defined for you.
# @param version, an integer
# @return a bool
# def isBadVersion(version):

class Solution:
    def firstBadVersion(self, n):
        l, r = 1, n
        while l + 1 < r:
            mid = l + (r - l)//2
            if isBadVersion(mid):
                r = mid
            else:
                l = mid
        if isBadVersion(l):
            return l
        return r

 

先判断边界/corner case，然后再判断first index that is true。

# The isBadVersion API is already defined for you.
# @param version, an integer
# @return a bool
# def isBadVersion(version):

class Solution:
    def firstBadVersion(self, n):
        if isBadVersion(1) : return 1
        l, r = 1, n  # l must be false
        while l + 1 < r:
            mid = l + (r - l)//2
            if isBadVersion(mid):
                r = mid
            else:
                l = mid
        return r