[Leetcode]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.

每个相同的起始元素对应(n-1)!个permutation,只要当前的k进行(n-1)!取余，得到的数字就是当前剩余数组的index，如此就可以得到对应的元素, 递推直到数组中没有元素结束~

class Solution:    # @return a string    def getPermutation(self, n, k):        factorial = 1; res = ""        for i in xrange(2, n):            factorial *= i        num = [i for i in xrange(1, n + 1)]        k -= 1        for i in reversed(xrange(n)):            res += str(num[k / factorial])            del num[k / factorial]            if i > 0:                k %= factorial                factorial /= i        return res