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.

public class Solution {
     /**
	 * This is a trick problem. We need to use how a permutation is generated.<br>
	 * Let us consider, for example, the first digit of the permutation.<br>
	 * Number 1 on the first digit until (n - 1)! + 1-th permutation. and (n - 1)! is the last
	 * permutation starts with 1. For the same reason 2*(n-1)!  is the last permutation starts with
	 * number 2 and so on.  So, on the first digit of k-th permutation can be determined by
	 * ((--k) / (n - 1)!). Recursively, we can find the other digits of the permutation.<br>
	 * 	 
	 * @param n --int, the number of digits in the permutation.
	 * @param k --int, the k-th permutation which is looking for.
	 * @return String --the permutation.
	 * @author Averill Zheng
	 * @version 2014-06-14
	 * @since JDK 1.7
	 */
	public String getPermutation(int n, int k) {
		List<Integer> num = new ArrayList<Integer>();
		int count = 1;
		//add the number in the ArrayList and calculate the factorial
		for(int i = 0; i < n; ++i){
			num.add((i + 1));
			count *= (i + 1);
		}
		
		//decrease k by 1
		--k;
		
		//find each digit of the permutation.
		StringBuffer stb = new StringBuffer();
		for(int i = 0; i < n; ++i){
			count /= (n - i);
			//select an item from the ArrayList and add to the permutation. After that remove 
			//the digit from the ArrayList since it has been added to the permutation.
			int itemToBeAdded = num.remove(k / count);
			stb.append(itemToBeAdded); 
			k %= count;
		}
		return stb.toString();
    }
}