[leetcode] 1492. The kth Factor of n

[leetcode] 1492. The kth Factor of n

Description

Given two positive integers n and k.

A factor of an integer n is defined as an integer i where n % i == 0.

Consider a list of all factors of n sorted in ascending order, return the kth factor in this list or return -1 if n has less than k factors.

Example 1:

Input: n = 12, k = 3Output: 3Explanation: Factors list is [1, 2, 3, 4, 6, 12], the 3rd factor is 3.

Example 2:

Input: n = 7, k = 2Output: 7Explanation: Factors list is [1, 7], the 2nd factor is 7.

Example 3:

Input: n = 4, k = 4Output: -1Explanation: Factors list is [1, 2, 4], there is only 3 factors. We should return -1.

Example 4:

Input: n = 1, k = 1Output: 1Explanation: Factors list is [1], the 1st factor is 1.

Example 5:

Input: n = 1000, k = 3Output: 4Explanation: Factors list is [1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000].

Constraints:

1 <= k <= n <= 1000

分析

题目的意思是：求出n个整除的数，然后排序，返回第k个数。这道题思路很直接，先从1~sqrt(n)中找出所有整除的数，然后排序返回第k个数就行了，稍微暴力了一点。

我看其他人也有直接遍历，边遍历边统计是否第K个值的做法，时间复杂度O(n)；还有一个是基于我这种思路，但是没有进行排序，因为遍历求出的整除的数已经有先后顺序了，所以这也可以优化一下，当然有兴趣可以试试，也比较简单哈。

代码

class Solution: def kthFactor(self, n: int, k: int) -> int: res=[] for i in range(int(sqrt(n))): num=i+1 if(n%num==0): res.append(num) res.append(n//num) res=list(set(res)) res.sort() if(k>len(res)): return -1 return res[k-1]

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