[LeetCode] 367. Valid Perfect Square

Given a positive integer num, write a function which returns True if num is a perfect square else False.

Follow up: Do not use any built-in library function such as sqrt.

Example 1:

Input: num = 16
Output: true

Example 2:

Input: num = 14
Output: false

Constraints:

• 1 <= num <= 2^31 - 1

有效的完全平方数。

这题做法几乎跟69题一模一样，题意是判断给定的数字是否是一个完全平方数。因为给的input数字有可能很大所以在计算mid的时候需要用long型来处理，否则会超时。

时间O(logn)

空间O(1)

Java实现

class Solution {
    public boolean isPerfectSquare(int num) {
        // corner case
        if (num == 1) {
            return true;
        }

        // normal case
        int low = 1;
        int high = num;
        while (low <= high) {
            long mid = low + (high - low) / 2;
            if (mid * mid == num) {
                return true;
            } else if (mid * mid < num) {
                low = (int) mid + 1;
            } else {
                high = (int) mid - 1;
            }
        }
        return false;
    }
}

JavaScript实现

/**
 * @param {number} num
 * @return {boolean}
 */
var isPerfectSquare = function(num) {
    // corner case
    if (num <= 0) {
        return false;
    }
    if (num === 1) {
        return true;
    }

    // normal case
    let low = 1;
    let high = num;
    while (low <= high) {
        let mid = parseInt(low + (high - low) / 2);
        if (mid * mid === num) {
            return true;
        } else if (mid * mid < num) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return false;
};

LeetCode 题目总结