[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