[LC] 90. Subsets II

Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

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

Solution 1

class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        if (nums == null || nums.length == 0) {
            return res;
        }
        Arrays.sort(nums);
        helper(res, new ArrayList<>(), nums, 0);
        return res;
    }
    
    private void helper(List<List<Integer>> res, List<Integer> list, int[] nums, int index) {
        res.add(new ArrayList<>(list));
        for (int i = index; i < nums.length; i++) {
            if (i > index && nums[i] == nums[i - 1]) {
                continue;
            }
            list.add(nums[i]);
            // use i not index b/c index smaller than i
            helper(res, list, nums, i + 1);
            list.remove(list.size() - 1);
        }
    }
}

class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        if (nums == null) {
            return result;
        }
        List<Integer> list = new ArrayList<>();
        Arrays.sort(nums);
        helper(nums, list, 0, result);
        return result;
    }
    
    private void helper(int[] nums, List<Integer> list, int level, List<List<Integer>> result)     {
        if (level == nums.length) {
            result.add(new ArrayList<>(list));
            return;
        }
        // need to add first, skip index and then remove
        list.add(nums[level]);
        helper(nums, list, level + 1, result);
        list.remove(list.size() - 1);
        
        while (level + 1 < nums.length && nums[level] == nums[level + 1]) {
            level += 1;
        }
        helper(nums, list, level + 1, result);
    }
}