SubsetII

Given a collection of integers that might contain duplicates, S, 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 S = [1,2,2], a solution is:

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

public class Solution {    public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) {// Start typing your Java solution below// DO NOT write main() functionArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();res.add(new ArrayList<Integer>());Arrays.sort(num);int start = 0;for(int i = 0; i < num.length; i++){int size = res.size();for(int j = start; j < size; j++){ArrayList<Integer> sub = new ArrayList<Integer>(res.get(j));sub.add(num[i]);res.add(sub);}if(i < num.length - 1 && num[i + 1] == num[i])start = size;elsestart = 0;}return res;}}

For [1, 2, 3], order:

[]

[1]

[2] [1, 2]

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

For [1, 2, 2], order:

[]

[1]

[2] [1, 2]

(start = size) [2, 2] [1, 2, 2]

public class Solution {    ArrayList<ArrayList<Integer>> ret;ArrayList<Integer> cur;public void DFS(int[] num, int n) {if (num.length == n) {ret.add(new ArrayList<Integer>(cur));return;}int count = 1;while (n + 1 < num.length && num[n] == num[n + 1]) {count++;n++;}for (int i = 0; i <= count; i++) {for (int j = 0; j < i; j++)cur.add(num[n]);DFS(num, n + 1);for (int j = 0; j < i; j++)cur.remove(cur.size() - 1);}}public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) {// Start typing your Java solution below// DO NOT write main() functionret = new ArrayList<ArrayList<Integer>>();cur = new ArrayList<Integer>();Arrays.sort(num);DFS(num, 0);return ret;}}

For [1, 2, 3], order: [[], [3], [2], [2, 3], [1], [1, 3], [1, 2], [1, 2, 3]]