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道題 的遞歸 非遞歸 都會超時。

只能用數學的方法計算。

 1 class Solution(object):
 2 
 3     def getPermutation(self, n, k):
 4         """
 5         :type nums: List[int]
 6         :rtype: List[List[int]]
 7         """
 8         k = k -1
 9         nums = list(range(1, n + 1))
10         res = ‘‘
11         while
 
 len(nums)!=0:
12             ai =int(k/self.fn(n-1))
13             res += str(nums[ai])
14             del nums[ai]
15             k = k % self.fn(n-1)
16             n = n - 1
17         return res
18     def fn(self, n):
19         res = 1
20         while n != 0:
21             res = res * n
22             n = n - 1
23
 
         return res
24             

60. Permutation Sequence（求全排列的第k個排列）