【LeetCode】【Math】the kth factor of n

题目：

给定两个正整数n和k。

整数n的因数定义为整数i，其中n％i == 0。

考虑按升序排列的所有n个因子的列表，返回此列表中的第k个因子；如果n小于k个因子，则返回-1。

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

【解法】

class Solution:
    def kthFactor(self, n: int, k: int) -> int:
        for factor in range(1, n + 1):
            if n % factor == 0:
                k -= 1
                if k == 0:
                    return factor
        return -1

Runtime: 32 ms, faster than 82.00% of Python3 online submissions for The kth Factor of n.

Memory Usage: 14 MB, less than 25.00% of Python3 online submissions for The kth Factor of n.

k -= 1, if k ==0，通过这个方法取特定位置的元素、

import math
class Solution:
    def kthFactor(self, n: int, k: int) -> int:
        factors = []
        for i in range(1, math.isqrt(n) + 1):
            if n % i == 0:
                factors.append(i)
                k -= 1
            if k == 0: return i
        if factors[-1] ** 2 == n: factors.pop() #删除重复元素
        if k > len(factors): return -1
        return n // factors[-k]
        

 

Runtime: 44 ms, faster than 48.07% of Python3 online submissions for The kth Factor of n.

Memory Usage: 13.5 MB, less than 100.00% of Python3 online submissions for The kth Factor of n.