Permutations II

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

Example:

Input: [1,1,2]
Output:
[
  [1,1,2],
  [1,2,1],
  [2,1,1]
]

class Solution {
    public List<List<Integer>> permuteUnique(int[] nums) {
        List<List<Integer>> list = new ArrayList<>();
        boolean[] used = new boolean[nums.length];
        Arrays.sort(nums);
        backtrack(list, new ArrayList<>(), nums, used);
        return list;
    }

    private void backtrack(List<List<Integer>> list, List<Integer> tempList, int[] nums, boolean used[]) {
        if (tempList.size() == nums.length) {
            list.add(new ArrayList<>(tempList));
        } else {
            for (int i = 0; i < nums.length; i++) {
                // If this element is already used OR previous similar element was not used
                // ,then we will not use this one aswell.
                // And ofcourse this array has been sorted for nums(i)==nums(i-1) to work
                if (used[i] || i > 0 && nums[i] == nums[i - 1] && !used[i - 1])
                    continue;
                used[i] = true;
                tempList.add(nums[i]);
                backtrack(list, tempList, nums, used);
                tempList.remove(tempList.size() - 1);
                used[i] = false;
            }
        }
    }
}