LeetCode 374: Guess Number Higher or Lower

最新推荐文章于 2024-10-21 14:58:17 发布

Cycle1413

最新推荐文章于 2024-10-21 14:58:17 发布

阅读量380

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode 文章标签： leetcode C++

本文链接：https://blog.youkuaiyun.com/Cycle1413/article/details/51917821

LeetCode 专栏收录该内容

32 篇文章

订阅专栏

本文介绍了一种简单而有效的猜数字游戏算法——二分查找法。通过预定义的API反馈来判断所猜数字与目标数字的高低关系，最终确定正确答案。文章详细解释了二分查找的具体实现过程，并注意到了在计算中间值时避免整数溢出的问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

374. Guess Number Higher or Lower

Difficulty: Easy
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.

思路

二分查找。
需要注意数据类型，mid = (lower + higher) / 2时，lower + higher可能超出取值范围，数据类型不能为int，若非要用int，则应用以下方式：mid = lower + (higher - lower) / 2。

代码

[C++]

class Solution {
public:
    int guessNumber(int n) {
        int lower = 1;
        int higher = n;
        while (lower <= higher) {
            int mid = lower + (higher - lower) / 2;
            int res = guess(mid);
            if (res == 0)
                return mid;
            else if (res == 1)
                lower = mid + 1;
            else
                higher = mid - 1;
        }
        return -1;
    }
};