Leetcode 1362. Closest Divisors

Posted 2021-08-21 SnailTyan

文章作者：Tyan
博客：noahsnail.com  |  CSDN  |  简书

1. Description

2. Solution

**解析：**Version 1，绝对差最小的两个数是一个数的开平方，遍历过程中，没必要遍历所有的数，遍历到数的开方就已经遍历了所有的乘积组合。优先选择n+1的解。

• Version 1

class Solution:
    def closestDivisors(self, num: int) -> List[int]:
        y1 = num + 1
        y2 = num + 2
        result = [0, 0]
        for i in range(1, int(sqrt(y2)) + 1):
            if y1 % i == 0:
                result[0] = i
                result[1] = y1 // i
            elif y2 % i == 0:
                result[0] = i
                result[1] = y2 // i
        return result

Reference

1. https://leetcode.com/problems/closest-divisors/