374. 猜数字大小

🟢   🔖  二分查找 交互  🔗 力扣 LeetCode

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 will tell you whether the number I picked is higher or lower than your guess.

You call a pre-defined API int guess(int num), which returns three possible results:

• -1: Your guess is higher than the number I picked (i.e. num > pick).
• 1: Your guess is lower than the number I picked (i.e. num < pick).
• 0: your guess is equal to the number I picked (i.e. num == pick).

Return the number that I picked.

Example 1:

Input: n = 10, pick = 6

Output: 6

Example 2:

Input: n = 1, pick = 1

Output: 1

Example 3:

Input: n = 2, pick = 1

Output: 1

Constraints:

• 1 <= n <= 2^31 - 1
• 1 <= pick <= n

题目大意

猜数字游戏的规则如下：

• 每轮游戏，我都会从 1 到 n 随机选择一个数字。 请你猜选出的是哪个数字。
• 如果你猜错了，我会告诉你，你猜测的数字比我选出的数字是大了还是小了。

你可以通过调用一个预先定义好的接口 int guess(int num) 来获取猜测结果，返回值一共有 3 种可能的情况（-1，1 或 0）：

• -1：我选出的数字比你猜的数字小 pick < num
• 1：我选出的数字比你猜的数字大 pick > num
• 0：我选出的数字和你猜的数字一样。恭喜！你猜对了！pick == num

返回我选出的数字。

解题思路

可以采用二分查找解决。

复杂度分析

• 时间复杂度： O(logn)，二分查找的时间复杂度是 O(logn)。
• 空间复杂度： O(1)。

代码

/**
 * @param {number} n
 * @return {number}
 */
var guessNumber = function (n) {
	let left = 1;
	let right = n;
	while (left <= right) {
		let mid = Math.floor((left + right) / 2);
		if (guess(mid) == 0) {
			return mid;
		} else if (guess(mid) == 1) {
			left = mid + 1;
		} else if (guess(mid) == -1) {
			right = mid - 1;
		}
	}
};

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