60. Permutation Sequence（求全排列的第k个排列）

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.

 

如果用前面2道题 的递归 非递归 都会超时。

只能用数学的方法计算。

class Solution(object):

    def getPermutation(self, n, k):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        k = k -1
        nums = list(range(1, n + 1))
        res = ''
        while len(nums)!=0:
            ai =int(k/self.fn(n-1))
            res += str(nums[ai])
            del nums[ai]
            k = k % self.fn(n-1)
            n = n - 1
        return res
    def fn(self, n):
        res = 1
        while n != 0:
            res = res * n
            n = n - 1
        return res