LeetCode 374. Guess Number Higher or Lower

最新推荐文章于 2025-09-11 18:44:42 发布

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最新推荐文章于 2025-09-11 18:44:42 发布

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文章标签：数据结构与算法 java

原文链接：http://www.cnblogs.com/Dylan-Java-NYC/p/6351273.html

本文介绍了一个经典的猜数字游戏的LeetCode题解。通过使用二分查找算法，可以在O(logn)的时间复杂度内找到正确答案。文章详细解释了如何通过预定义的API guess(int num)来判断猜测的数字与目标数字之间的大小关系，并据此调整搜索范围。

原题链接在这里：https://leetcode.com/problems/guess-number-higher-or-lower/

题目：

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.

题解：

Binary Search, guess mid为0则return mid. guess mid为-1, 就在左半面继续猜. guess mid 为1, 就在右半面继续猜.

Time Complexity: O(logn). Space: O(1).

AC Java:

 1 /* The guess API is defined in the parent class GuessGame.
 2    @param num, your guess
 3    @return -1 if my number is lower, 1 if my number is higher, otherwise return 0
 4       int guess(int num); */
 5 
 6 public class Solution extends GuessGame {
 7     public int guessNumber(int n) {
 8         int l = 1;
 9         int r = n;
10         while(l <= r){
11             int mid = l + (r-l)/2;
12             int result = guess(mid);
13             if(result < 0){
14                 r = mid-1;
15             }else if(result > 0){
16                 l = mid+1;
17             }else{
18                 return mid;
19             }
20         }
21         return -1;
22     }
23 }