60. Permutation Sequence（求全排列的第k个排列）

The set [1,2,3,…,n] contains a total of n! unique permutations.

By listing and labeling all of the permutations in order,
We get the following sequence (ie, for n = 3):

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

Given n and k, return the kth permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

如果用前面2道题 的递归 非递归 都会超时。

只能用数学的方法计算。

 class Solution(object):

     def getPermutation(self, n, k):
         """
         :type nums: List[int]
         :rtype: List[List[int]]
         """
         k = k -1
         nums = list(range(1, n + 1))
         res = ''
         while len(nums)!=0:
             ai =int(k/self.fn(n-1))
             res += str(nums[ai])
             del nums[ai]
             k = k % self.fn(n-1)
             n = n - 1
         return res
     def fn(self, n):
         res = 1
         while n != 0:
             res = res * n
             n = n - 1
         return res