278. 第一个错误的版本

题目描述

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通过质量检测。由于每个版本都是基于之前的版本开发的，所以错误的版本之后的所有版本都是错的。

假设你有 n 个版本 [1, 2, ..., n]，你想找出导致之后所有版本出错的第一个错误的版本。

你可以通过调用 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。你应该尽量减少对调用 API 的次数。

示例 1：

输入：n = 5, bad = 4
输出：4
解释：
调用 isBadVersion(3) -> false 
调用 isBadVersion(5) -> true 
调用 isBadVersion(4) -> true
所以，4 是第一个错误的版本。

示例 2：

输入：n = 1, bad = 1
输出：1

提示：

1 <= bad <= n <= 2³¹ - 1

解法

方法一

Python3JavaC++GoRustJavaScript

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


class Solution:
    def firstBadVersion(self, n):
        """
        :type n: int
        :rtype: int
        """
        left, right = 1, n
        while left < right:
            mid = (left + right) >> 1
            if isBadVersion(mid):
                right = mid
            else:
                left = mid + 1
        return left

/* The isBadVersion API is defined in the parent class VersionControl.
      boolean isBadVersion(int version); */

public class Solution extends VersionControl {
    public int firstBadVersion(int n) {
        int left = 1, right = n;
        while (left < right) {
            int mid = (left + right) >>> 1;
            if (isBadVersion(mid)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

// The API isBadVersion is defined for you.
// bool isBadVersion(int version);

class Solution {
public:
    int firstBadVersion(int n) {
        int left = 1, right = n;
        while (left < right) {
            int mid = left + ((right - left) >> 1);
            if (isBadVersion(mid)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
};

/**
 * Forward declaration of isBadVersion API.
 * @param   version   your guess about first bad version
 * @return            true if current version is bad
 *                    false if current version is good
 * func isBadVersion(version int) bool;
 */

func firstBadVersion(n int) int {
    left, right := 1, n
    for left < right {
        mid := (left + right) >> 1
        if isBadVersion(mid) {
            right = mid
        } else {
            left = mid + 1
        }
    }
    return left
}

// The API isBadVersion is defined for you.
// isBadVersion(version:i32)-> bool;
// to call it use self.isBadVersion(version)

impl Solution {
    pub fn first_bad_version(&self, n: i32) -> i32 {
        let mut left = 1;
        let mut right = n;
        while left < right {
            let mid = left + (right - left) / 2;
            if self.isBadVersion(mid) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        left
    }
}

/**
 * Definition for isBadVersion()
 *
 * @param {integer} version number
 * @return {boolean} whether the version is bad
 * isBadVersion = function(version) {
 *     ...
 * };
 */

/**
 * @param {function} isBadVersion()
 * @return {function}
 */
var solution = function (isBadVersion) {
    /**
     * @param {integer} n Total versions
     * @return {integer} The first bad version
     */
    return function (n) {
        let left = 1;
        let right = n;
        while (left < right) {
            const mid = (left + right) >>> 1;
            if (isBadVersion(mid)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    };
};

278. 第一个错误的版本

题目描述

解法

方法一

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