【leetcode刷题】60 Permutation Sequence (python)

原题链接
https://leetcode.com/problems/permutation-sequence/
解题思路
每次用k来除以(n-1)!，k更新为k%(n-1)!，得到第一个数字的序号；
然后在数列中去掉这个数字，继续按照这种方法确定第二个数字在数列中的序号，挨着确定后面每一位的数字。

代码

class Solution(object):
    def getPermutation(self, n, k):
        """
        :type n: int
        :type k: int
        :rtype: str
        """  
        nums = [i+1 for i in range(n)]
        return self.count(n, k, nums, '')

    def count(self, n, k, nums, string):
        if k==0:
            string += ''.join(str(item) for item in nums[::-1])
            return string
        divisor = 1
        for i in range(n-1, 0, -1):
            divisor *= i
        if (k%divisor) != 0:
            tmp = k // divisor
        else:
            tmp = k // divisor - 1
        k = k % divisor
        sub_nums = nums[:]
        del sub_nums[tmp]
        return self.count(n-1, k, sub_nums, string+str(nums[tmp]))