LeetCode278. First Bad Version-优快云博客

本文介绍了一种使用二分查找算法来高效确定首次出现故障的产品版本的方法。在一个包含多个连续开发的软件版本中，若某一版本及其后续版本均存在故障，通过调用isBadVersion API，本文展示了如何快速定位到首个故障版本，以最小化API调用次数。

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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 versionis 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.

思路:这个题的意思就是找到第一个返回为True的version.最开始的想法肯定是从后向前遍历,当返回值出现第一个false时(位置为n),返回n+1.但这种方法效率很低O(n),在leetcode中会显示超时.所以需要改进,用二分查找的方式会降低时间复杂度O(log2n).

需要注意的是和二分查找不同,right不能等于mid-1,因为这样可能会漏掉第一个true的位置.

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

class Solution:
    def firstBadVersion(self, n):
        """
        :type n: int
        :rtype: int
        """    
        left,right = 1,n

        while left<right:  
            mid = (left+right)//2
            if isBadVersion(mid):
                right = mid
            else:
                left = mid + 1
        
        return left