60. Permutation Sequence 【M】【12】

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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比较麻烦的地方在于，当整除的时候。。这个就需要特殊处理一下

class Solution(object):    def getPermutation(self, n, k):                res = ''        nums = [0] * n        for i in xrange(n):            nums[i] = i + 1                while k > 1:                        f = math.factorial(n-1)            if k % f == 0:                k -= 1                pos = k/f                res += str(nums[pos])                del(nums[pos])                nums = nums[::-1]                break                            #print k,f,k/f                        pos = k/f            res += str(nums[pos])                        del(nums[pos])                        k %= f            n -= 1                for i in nums:            res += str(i)        return res