90. Subsets II

Given a collection of integers that might contain duplicates, nums, return all possible subsets.

Note: The solution set must not contain duplicate subsets.

For example,
If nums = [1,2,2], a solution is:

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

Solution:

We use DFS to traverse all possible subsets.

The tricky part is to remove duplicates.

So we first sort the array.

During traversing, when we find a num[i] == num[i - 1], and num[i - 1] has not been selected, we will not select num[i - 1].



Code:

public class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        if (nums == null || nums.length == 0) {
            return result;
        }
        Arrays.sort(nums);
        List<Integer> list = new ArrayList<>();
        helper(nums, result, list, 0);
        return result;
    }
    
    public void helper(int[] nums, List<List<Integer>> result, List<Integer> list, int pos) {
        result.add(new ArrayList<Integer>(list));
        
        for (int i = pos; i < nums.length; i++) {
            if (i > pos && nums[i] == nums[i - 1]) {
                continue;
            }
            list.add(nums[i]);
            helper(nums, result, list, i + 1);
            list.remove(list.size() - 1);
        }
    }
}