LeetCode | 0278. First Bad Version第一个错误的版本【Python】

LeetCode 0278. First Bad Version第一个错误的版本【Easy】【Python】【二分】

Problem

LeetCode

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.

问题

力扣

你是产品经理，目前正在带领一个团队开发新的产品。不幸的是，你的产品的最新版本没有经过质量检测。因为每一个版本都是基于以前的版本开发的，因此错误的版本以后的全部版本都是错的。

假设你有 n 个版本 [1, 2, ..., n]，你想找出致使以后全部版本出错的第一个错误的版本。

你能够经过调用 bool isBadVersion(version) 接口来判断版本号 version 是否在单元测试中出错。实现一个函数来查找第一个错误的版本。你应该尽可能减小对调用 API 的次数。

示例:

给定 n = 5，而且 version = 4 是第一个错误的版本。

调用 isBadVersion(3) -> false
调用 isBadVersion(5) -> true
调用 isBadVersion(4) -> true

因此，4 是第一个错误的版本。

思路

二分查找

由于版本是从 1 到 n，因此 low 初值设为 1，high 初值设为 n。

时间复杂度: O(logn)
空间复杂度: O(1)

Python代码

# 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
        """
        low, high = 1, n  # 1-n
        while low <= high:
            mid = int((low + high) / 2)
            if isBadVersion(mid) == False:
                low = mid + 1
            else:
                high = mid - 1
        return low

代码地址

GitHub连接