# Subsets

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],
[]
]
``````

Tips:

1. recursion 方法注意for循环里面， helper是i+1，比较难想；还有for里面i是从depth开始不是0！！
2. 大神方法就是注意各种new, 各种添加；O(n^2)
3. 位运算很精妙，建议找机会听个讲座if ((i & (1 << j)) != 0)

Code:

recursive:

``````public class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
if (nums == null || nums.length == 0) {
return result;
}
Arrays.sort(nums);
dfs(nums, result, new ArrayList<Integer>(), 0);
return result;
}
private void dfs(int[] nums, List<List<Integer>> result, List<Integer> path, int depth) {
for (int i = depth; i < nums.length; i++) {
dfs(nums, result, path, i + 1);
path.remove(path.size() - 1);
}
}
}
``````

``````class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
Arrays.sort(nums);
for (int i = 0; i < nums.length; i++) {
List<List<Integer>> newRes = new ArrayList<>(result);
for (List<Integer> l : result) {
ArrayList<Integer> newList = new ArrayList<>(l);
}
result = newRes;
}
return result;
}
}
``````

``````class Solution {
public ArrayList<ArrayList<Integer>> subsets(int[] nums) {
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
int n = nums.length;
Arrays.sort(nums);
for (int i = 1; i < (1 << n); i++) {
ArrayList<Integer> subset = new ArrayList<Integer>();
for (int j = 0; j < n; j++) {
if ((i & (1 << j)) != 0) {