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

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题目来源

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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):

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

Given n and k, return the kth permutation sequence.

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题意分析

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Input: n, k

Output: the kth permutation in permutations based on n

Conditions: 返回第n个排列

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题目思路

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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

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AC代码（Python）

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 class Solution(object):
     def getPermutation(self, n, k):
         """
         :type n: int
         :type k: int
         :rtype: str
         """
         res = ""

         nums = [i + 1 for i in xrange(n)]
         #count from 0
         k -= 1
         fac = 1
         for i in xrange(1, n):
             fac *= i
         for i in reversed(xrange(n)):
             index = k / fac
             value = nums[index]
             res += str(value)
             nums.remove(value)
             if i != 0:
                 k %= fac
                 fac /= i
         return res