[LeetCode]题解（python）：060-Permutation Sequence

题目来源

https://leetcode.com/problems/permutation-sequence/

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.

题意分析

Input: n, k

Output: the kth permutation in permutations based on n

Conditions: 返回第n个排列

题目思路

看了网上答案发现还有这样一个方法，每一位都是nums[k/factorial]，注意k要从0计数所以要减一，nums要及时移除已使用过的元素（不移除可以用False or True数组标记），factorial一般是n-1的阶乘，以上三个数都要更新

AC代码（Python）

 1 class Solution(object):
 2     def getPermutation(self, n, k):
 3         """
 4         :type n: int
 5         :type k: int
 6         :rtype: str
 7         """
 8         res = ""
 9         
10         nums = [i + 1 for i in xrange(n)]
11         #count from 0
12         k -= 1
13         fac = 1
14         for i in xrange(1, n):
15             fac *= i
16         for i in reversed(xrange(n)):
17             index = k / fac
18             value = nums[index]
19             res += str(value)
20             nums.remove(value)
21             if i != 0:
22                 k %= fac
23                 fac /= i
24         return res