DAY23：leetcode #60 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.

Note: Given n will be between 1 and 9 inclusive.

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class Solution(object):
    def getFactorial(self,n):
        if n == 1:
            self.factorial.append(1)
            return 1
        else:
            temp = n*self.getFactorial(n-1)
            self.factorial.append(temp)
            return temp
    def getPermutation(self, n, k):
        """
        :type n: int
        :type k: int
        :rtype: str
        """
        self.factorial = []
        self.getFactorial(n)
        str_list = []
        int_list = range(1,n+1)
        for f in self.factorial[:-1][::-1]:
            str_list.append(str(int_list[(k - 1)/f]))
            del int_list[(k - 1)/f]
            k = k%f
        str_list.append(str(int_list[0]))
        return ''.join(str_list)