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.

Subscribe to see which companies asked this question

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)