# Given a set of integers, return all possible permutations that contains duplicate.

Consider example `[1,2,2]` returns

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

This problem is classic example of backtracking. If there are duplicate elements, the total number of elements are n!/(k!z!) where n is the total number of elements and k,z are duplicate count for an integer.

``````public class Permutations {

public static void main(String[] args) {
System.out.println(new Permutations2().permuteUnique(new int[] { 2, 2, 1, 1 }));
}

public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> bigList =
new ArrayList<List<Integer>>();
Arrays.sort(nums);
permute(nums, 0, bigList);
return bigList;
}

private void permute(int[] nums, int index, List<List<Integer>> bigList) {
if (index == nums.length) {
List l = new ArrayList<Integer>(nums.length);
for (int num : nums)
return;
}
Set<Integer> dups = new HashSet();
for (int i = index; i < nums.length; i++) {
swap(nums, i, index);
permute(nums, index + 1, bigList);
swap(nums, i, index);
}
}
}

private void swap(int[] nums, int i, int index) {
int temp = nums[i];
nums[i] = nums[index];
nums[index] = temp;
}
}
``````