374. Guess Number Higher or Lower

最新推荐文章于 2024-10-21 14:58:17 发布

原创最新推荐文章于 2024-10-21 14:58:17 发布 · 194 阅读

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本文介绍了一种猜数字游戏的高效解决策略。通过分析预定义的API guess(int num)反馈，利用二分查找法实现快速猜测目标数字。文章详细解释了如何避免整数溢出问题，并提供了完整的代码实现。

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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.

这是一个查找问题，首先直接使用for循环查找，会超时，然后采用二分查找。

// 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) {
        int low = 1;
        while(low <= n)
        {
            int mid = (n + low)/2;
            int m = guess(mid);
            if(m == 0)
            return mid;
            else if(m == -1)
            n = mid -1;
            else
            low = mid + 1;
        }
        //return -1;
    }
};

提交后发现出现Time Limit Exceeded 2126753390 1702766719

这是因为其中mid的值溢出了，修改之后通过。

// 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) {
        int low = 1;
        while(low <= n)
        {
            int mid = (n - low)/2 + low;
            int m = guess(mid);
            if(m == 0)
            return mid;
            else if(m == -1)
            n = mid -1;
            else
            low = mid + 1;
        }
        //return -1;
    }
};