# 78 Subsets – Medium

### Problem:

Given a set of distinct integers, nums, return all possible subsets.

Note:

Elements in a subset must be in non-descending order. 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], [] ]

### Thoughts:

Init solution set with empty list.

Iterating over all numbers, for each number, add it to all existing subsets, but keep the existing subests.

Say for numbers [1, 2, 3],

Init solution set to be []

1: [], [1]

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

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

### Solutions:

``````public class Solution {
public List<List<Integer>> subsets(int[] nums) {
if (nums == null)
return null;
Arrays.sort(nums);
List<List<Integer>> result = new ArrayList<List<Integer>>();
List<List<Integer>> toAddAll = new ArrayList<List<Integer>>();
for (int i = 0; i < nums.length; i++) {
//get sets that are already in result
for (List<Integer> a : result) {
List<Integer> toAdd = new ArrayList<Integer>(a);
}
}
return result;
}
}
``````

Recursion version:

``````public class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> result = new LinkedList<List<Integer>>();
Arrays.sort(nums);
process(0, nums, result, new LinkedList<Integer>());
return result;
}
private void process(int start, int[] nums, List<List<Integer>> result, List<Integer> curr) {