LeetCode | 1362. Closest Divisors最接近的因数【Python】
LeetCode 1362. Closest Divisors最接近的因数【Medium】【Python】【数学】
Problem
Given an integer num
, find the closest two integers in absolute difference whose product equals num + 1
or num + 2
.
Return the two integers in any order.
Example 1:
Input: num = 8 Output: [3,3] Explanation: For num + 1 = 9, the closest divisors are 3 & 3, for num + 2 = 10, the closest divisors are 2 & 5, hence 3 & 3 is chosen.
Example 2:
Input: num = 123 Output: [5,25]
Example 3:
Input: num = 999 Output: [40,25]
Constraints:
1 <= num <= 10^9
问题
给你一个整数 num
,请你找出同时满足下面全部要求的两个整数:
- 两数乘积等于
num + 1
或num + 2
- 以绝对差进行度量,两数大小最接近
你可以按任意顺序返回这两个整数。
示例 1:
输入:num = 8 输出:[3,3] 解释:对于 num + 1 = 9,最接近的两个因数是 3 & 3;对于 num + 2 = 10, 最接近的两个因数是 2 & 5,因此返回 3 & 3 。
示例 2:
输入:num = 123 输出:[5,25]
示例 3:
输入:num = 999 输出:[40,25]
提示:
1 <= num <= 10^9
思路
数学
从 1 遍历到 sqrt(x),找到最接近的两个因数。 其实从 sqrt(x) 倒过来遍历会更快。
时间复杂度: O(sqrt(n))
空间复杂度: O(1)
Python3代码
class Solution: def closestDivisors(self, num: int) -> List[int]: import math num1, num2 = num + 1, num + 2 ans1 = self.crack(num1) ans2 = self.crack(num2) res = [] if abs(ans1 - int(num1 / ans1)) < abs(ans2 - int(num2 / ans2)): # int res.append(ans1) res.append(int(num1 / ans1)) return res else: res.append(ans2) res.append(int(num2 / ans2)) return res # calculate factor def crack(self, integer): factor1, factor2 = 1, integer for i in range(1, int(math.sqrt(integer)) + 1): # range: [1, sqrt(x) + 1) if int(integer / i) == integer / i: # i is factor if integer / i - i < factor2 - factor1: factor1, factor2 = i, integer / i return factor1