Leetcode Guess Number Higher or Lower 374

最新推荐文章于 2025-08-19 21:22:03 发布

原创最新推荐文章于 2025-08-19 21:22:03 发布 · 390 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

算法练习专栏收录该内容

141 篇文章

订阅专栏

本文介绍了一种猜数字游戏的算法实现，通过折半查找的方法高效地从1到n中找出预设的目标数字。该算法利用预先定义好的API来获取每次猜测的反馈，并据此调整搜索范围，最终确定目标数字。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

We are playing the Guess Game. The game is as follows:

I pick a number from 1 to n. You have to guess which number I picked.

Every time you guess wrong, I’ll tell you whether the number is higher or lower.

You call a pre-defined API guess(int num) which returns 3 possible results (-1, 1, or 0):

-1 : My number is lower
1 : My number is higher
0 : Congrats! You got it!
Example:
n = 10, I pick 6.

Return 6.

题目链接

折半查找，注意边界，另外题目给定一定会找到
另外，使用无符号整数

// Forward declaration of guess API.
// @param num, your guess
// @return -1 if my number is lower, 1 if my number is higher, otherwise return 0
int guess(int num);

class Solution {
public:
    int guessNumber(int n) {
        unsigned left=0,right=n,mid;
        while(left<=right){
            mid=(left+right)/2;
            //cout<<mid<<endl;
            if(temp_ans==0) break;
            else if(temp_ans==1) left=mid+1;
            else right=mid-1;
        }
        return mid;
    }
};