LeetCode | 0278. First Bad Version第一個錯誤的版本【Python】

LeetCode 0278. First Bad Version第一個錯誤的版本【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.

問題

力扣

你是產品經理，目前正在帶領一個團隊開發新的產品。不幸的是，你的產品的最新版本沒有經過質量檢測。因爲每一個版本都是基於以前的版本開發的，因此錯誤的版本以後的全部版本都是錯的。

假設你有 n 個版本 [1, 2, ..., n]，你想找出致使以後全部版本出錯的第一個錯誤的版本。

你能夠經過調用 bool isBadVersion(version) 接口來判斷版本號 version 是否在單元測試中出錯。實現一個函數來查找第一個錯誤的版本。你應該儘可能減小對調用 API 的次數。

示例:

給定 n = 5，而且 version = 4 是第一個錯誤的版本。

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

因此，4 是第一個錯誤的版本。

思路

二分查找

由於版本是從 1 到 n，因此 low 初值設爲 1，high 初值設爲 n。

時間複雜度: O(logn)
空間複雜度: O(1)

Python代碼

# 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

代碼地址

GitHub連接