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 {
	int f[];
    public String getPermutation(int n, int k) {
    	f = new int[n+1];
    	if(k>factorial(n)) return "-1";
    	List<Integer> list = new ArrayList<Integer>();
        for(int i=1;i<=n;i++){
        	list.add(i);
        }
        StringBuilder sb = new StringBuilder();
        k--;
        while(list.size()>0){
        	int mul = factorial(n-1);
        	int index = k/mul;
        	sb.append(list.get(index));
        	list.remove(index);
        	k = k%mul;
        	n--;
        }
        return sb.toString();
    }
    private int factorial(int num){
    	if(num == 0) return f[0]=1;
    	if(f[num]>0){
    		return f[num];
    	}else{
    		return f[num]=factorial(num-1)*num;
    	}
    	
    }
}

[leetCode]Permutation Sequence