Leetcode 278 第一个错误的版本 ( First Bad Version *Easy* ) 题解分析

题目介绍

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 returns whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.

示例

Example 1:

Input: n = 5, bad = 4
Output: 4
Explanation:
call isBadVersion(3) -> false
call isBadVersion(5) -> true
call isBadVersion(4) -> true
Then 4 is the first bad version.

Example 2:

Input: n = 1, bad = 1
Output: 1

简析

简单来说就是一个二分查找，但是这个问题其实处理起来还是需要搞清楚一些边界问题

代码

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public int firstBadVersion(int n) {
    // 类似于双指针法
    int left = 1, right = n, mid;
    while (left < right) {
        // 取中点
        mid = left + (right - left) / 2;
        // 如果不是错误版本，就往右找
        if (!isBadVersion(mid)) {
            left = mid + 1;
        } else {
        // 如果是的话就往左查找
            right = mid;
        }
    }
    // 这里考虑交界情况是，在上面循环中如果 left 是好的，right 是坏的，那进入循环的时候 mid == left
    // 然后 left = mid + 1 就会等于 right，循环条件就跳出了，此时 left 就是那个起始的错误点了
    // 其实这两个是同一个值
    return left;
}

往右移动示例

往左移动示例

结果