Subsets II

Given a list of numbers that may has duplicate numbers, return all possible subsets

Example

If S = [1,2,2], a solution is:

[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

Note

• Each element in a subset must be in non-descending order.
• The ordering between two subsets is free.
• The solution set must not contain duplicate subsets.

Challenge

Can you do it in both recursively and iteratively?

public class Solution {
    public ArrayList<ArrayList<Integer>> subsetsWithDup(ArrayList<Integer> S) {
         ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
         ArrayList<Integer> list = new ArrayList<Integer>();
         if (S == null || S.size() == 0) {
             return result;
         }
         Collections.sort(S);
         subsetsHelper(result, list, S, 0);
         return result;
    }
    private void subsetsHelper(ArrayList<ArrayList<Integer>> result, ArrayList<Integer> list, ArrayList<Integer> S, int pos) {
        result.add(new ArrayList<Integer>(list));
        for (int i = pos; i < S.size(); i++) {
            if (i != pos && S.get(i) == S.get(i - 1)) {
                continue;
            }    
            list.add(S.get(i));
            subsetsHelper(result, list, S, i + 1);
            list.remove(list.size() - 1);
        }
    }
}