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):
"123""132""213""231""312""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。
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
