Permutation sequence

This is from leetcode:

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):

"123"
"132"
"213"
"231"
"312"
"321"

Given n and k, return the kth permutation sequence. (Note: Given n will be between 1 and 9 inclusive.)

The best way to approach the problem is to divide and conquer:
let's start with numbers 1, 2, 3, 4, and we can divide the sequences into 4 groups each starts with 1, 2, 3, 4, the number of sequences in each are 3!,  k/3! will decide which group the kth sequence belongs to, saying k=9, k/3! = 1, so kth sequence belongs with group starts with 2.

now we have 1,3,4 and we should look for k%3!=9%6=3th sequence in this set

And we can use recursion to do divide and conquer work.

public static List<Integer> permutationByRank(int n, int k){
       //prepare the set
       List<Integer> list = new ArrayList<Integer>();
        for(int i=1; i<=n; i++)
            list.add(i);
        //calculate the group size
        int size = factor(n-1);
        return permuByRankUtil(list, k-1, size);
    }
   
    /*
     * here k starts at 0 to n!-1 for easy computing purpose
     */
    public static List<Integer> permuByRankUtil(List<Integer> nums, int k, int f){
        int size = nums.size();
        //base case
        if(size==1)
            return nums;
        else{
            //decide group index
            int index = k/f;
            int first = nums.remove(index);
            List<Integer> re = permuByRankUtil(nums, k%f, f/(size-1));
            re.add(0, first);
            return re;
        }
    }