leetcode-6-猜数字大小

最新推荐文章于 2022-04-26 09:27:12 发布

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本文详细解析了猜数字游戏的算法实现，通过二分查找法在1到n之间找到目标数字。介绍了预定义API guess(int num)的使用，以及如何处理大数问题和C++取整特性。附带C++代码示例。

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Guess my number

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 :

Input: n = 10, pick = 6
Output: 6

这个题虽然简单但也用了半个小时，还看了一下评论，有两个半坑：

这里的my number是指他的数字更大一些或小一些，所以看英文的话会更好理解题意
C++取整是向下取整，所以如果pick是n的话永远取不到，为了不用round四舍五入函数，所以要从n开始
有非常大的数，int表示不了

// 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) {
        long int low = 1;
        long int high = n;
        long int res = high;//因为是向下取整
        long int tof = guess(res);
        while(tof != 0)
        {
            if(tof == 0)
            {
                break;
            }
            else if(tof == 1)
            {
                low = res;
                res = (low+high)/2;
                tof = guess(res);
            }
            else
            {
                high = res;
                res = (low+high)/2;
                tof = guess(res);
            }
        }
        return res;
    }
};