【LeetCode】278. First Bad Version 解题报告（Python & C++）

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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目录

• • 题目描述
  • 解题方法
  • • 二分查找
  • 日期

题目地址：https://leetcode.com/problems/first-bad-version/description/

题目描述

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.

解题方法

二分查找

找出最开始错的版本。

其实就是二分查找，注意题目求的是版本号，而不是调用了多少次isBadVersion函数。

# 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
        """
        left = 0
        right = n - 1
        while left <= right:
            mid = left + (right - left) / 2
            if isBadVersion(mid):
                right = mid - 1
            else:
                left = mid + 1
        return left

C++版本：

// Forward declaration of isBadVersion API.
bool isBadVersion(int version);
class Solution {
public:
    int firstBadVersion(int n) {
        int left = 0, right = n; //[left, right]
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (isBadVersion(mid)) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
};

日期

2018 年 2 月 4 日
2018 年 11 月 28 日 —— 心无旁骛