LeetCode: Sum of Square Numbers

Given a non-negative integer c, your task is to decide whether there're two integers a and b such that a2 + b2 = c.

Example 1:

Input: 5
Output: True
Explanation: 1 * 1 + 2 * 2 = 5

方法一：

簡單粗暴，嵌套依次遍歷, 時間復(fù)雜度O(n^2)

I/O : 2147483641 // => (測試超時)

Detail -> Time Limit Exceeded

var judgeSquareSum = function(c) {
    let sq = Math.sqrt(c)
    
    for(let i = 0; i <= sq; i++) {
        for(let j = i; j <= sq; j++) {
            let pdt = i ** 2 + j ** 2
            if(pdt === c) {
                return true
            } else if(pdt > c) {
                break
            }
        }
    }
    
    return false
};

方法二：

二分法，先取 0，sqrt(c) ,在逐步往中間縮小范圍，時間復(fù)雜度 O(n)

I/O : 2147483641 // => false

Detail -> cases: 124, Runtime: 68 ms

var judgeSquareSum = function(c) {
    let b = Math.sqrt(c) | 0
    let a = 0
    
    while(a <= b) {
        if (a * a + b * b < c) {
            a--
        } else if (a * a + b * b === c) {
            return true
        } else {
            b--
        }
    }
    
    return false
};