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

答案
Permutation II 只需要在Permutation的基础上，确保同一个位置i上如果已经选择了数字num, 则不能再次选择(use a set for this)

class Solution {
    public List<List<Integer>> permuteUnique(int[] nums) {
        List<List<Integer>> list = new ArrayList<List<Integer>>();
        List<Integer> curr = new ArrayList<Integer>();
        boolean[] used = new boolean[nums.length];
        recur(nums, used, 0, curr, list);
        return list;
    }
    
    private void recur(int[] nums, boolean[] used, int i, List<Integer> curr, List<List<Integer>> list) {
        if(i == nums.length) {
            list.add(new ArrayList<Integer>(curr));
            return;
        }

        Set set = new HashSet<Integer>();
        for(int j = 0; j < nums.length; j++) {
            if(set.contains(nums[j])) continue;
            if(!used[j]) {
                set.add(nums[j]);
                curr.add(nums[j]);
                used[j] = true;
                recur(nums, used, i + 1, curr, list);
                used[j] = false;
                curr.remove(curr.size() - 1);                
            }
        }
    }
}