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.

原题链接：https://oj.leetcode.com/problems/permutation-sequence/

从n个数的全排列中找出第k个排列。

n开头的排列有(n-1)!个。

k/(n-1)!可确定第一个数字，在余下的(n-1)!中找k%(n-1)!个。

public class PermutationSequence {
	public String getPermutation(int n, int k) {
		List<Integer> list = new ArrayList<Integer>();
		for(int i=1;i<=n;i++)
			list.add(i);
		int mod = 1;
		for (int i = 1; i <= n; i++) {
			mod = mod * i;
		}
		k--;
		StringBuilder builder = new StringBuilder();
		for(int i=0;i<n;i++){
			mod /= (n-i);
			int index = k / mod;
			k %= mod;
			builder.append(list.get(index));
			list.remove(index);
		}
		return builder.toString();
	}
}