SubSets, also a basic method of permutation

Given a set of distinct integers, nums, return all possible subsets.
Note: The solution set must not contain duplicate subsets.
For example,
If nums = [1,2,3], a solution is:

[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
] 

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public class Solution { public List<List<Integer>> subsets(int[] nums) { if (nums == null){ return null; } //not necessary // Arrays.sort(nums); //this is also basic method of permutation List<List<Integer>> result = new ArrayList<List<Integer>>(); for (int i = 0; i < nums.length; i++) { //new list is created based on existing list, by adding one more new number List<List<Integer>> newList = new ArrayList<List<Integer>>(); //build the newlist from existing list, existing lists are still there for (List<Integer> a : result) { newList.add(new ArrayList<Integer>(a)); } //add new number to existing sets for (List<Integer> a : newList) { a.add(nums[i]); } //in a loop of each element, creating single element set List<Integer> single = new ArrayList<Integer>(); single.add(nums[i]); newList.add(single); //adding new lists to result, when result is processed in next loop, more content will be added the ///same way result.addAll(newList); } //add empty set. it will be wrong if adding empty set the first, because it will be permunated with //other new elements result.add(new ArrayList<Integer>()); return result; } }