Given a collection of numbers that might contain duplicates, return all possible unique permutations.For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].

1         if (permutation.size() == num.length) {
2             if (!permutations.contains(permutation)) {
3                 permutations.add(new ArrayList<Integer>(permutation));
4                 return;
5             }
6         }    

 1 public class Solution {
 2     public ArrayList<ArrayList<Integer>> permuteUnique(int[] num) {
 3         Arrays.sort(num);
 4         ArrayList<Integer> permutation = new ArrayList<Integer>();
 5         ArrayList<ArrayList<Integer>> permutations = new ArrayList<ArrayList<Integer>>();
 6         boolean[] visited = new boolean[num.length];
 7         helper(permutation, permutations, num, visited);
 8         return permutations;
 9     }
10
11     public void helper(ArrayList<Integer> permutation, ArrayList<ArrayList<Integer>> permutations, int[] num, boolean[] visited) {
12         if (permutation.size() == num.length) {
13                 permutations.add(new ArrayList<Integer>(permutation));
14                 return;
15         }
16
17         for (int k = 0; k < num.length; k++) {
18             if (k > 0 && !visited[k-1] && num[k] == num[k-1]) continue;
19             if (!visited[k]) {
20                 visited[k] = true;
21                 permutation.add(num[k]);
22                 helper(permutation, permutations, num, visited);
23                 permutation.remove(permutation.size() - 1);
24                 visited[k] = false;
25             }
26         }
27     }
28 }