【leetcode】1492. The kth Factor of n

题目如下：

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

Example 2:

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

Example 3:

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

Example 4:

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

Example 5:

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

Constraints:

• 1 <= k <= n <= 1000

解题思路：n最大才1000，把n全部的因子求出来排序就可以了。

代码如下：

class Solution(object):
    def kthFactor(self, n, k):
        """
        :type n: int
        :type k: int
        :rtype: int
        """
        factor = []
        for i in range(1,n+1):
            if n%i == 0:
                factor.append(i)
                if n/i != i : factor.append(n/i)
        factor = sorted(set(factor))
        return factor[k-1] if k <= len(factor) else -1