lintcode18- subsetsII- medium

Given a list of numbers that may has duplicate numbers, return all possible subsets

 Notice

• Each element in a subset must be in non-descendingorder.
• The ordering between two subsets is free.
• The solution set must not contain duplicate subsets.

Example

If S = [1,2,2], a solution is:

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

Challenge

Can you do it in both recursively and iteratively?

和 I 一样，仅仅加一个判断if，if (i == index || nums[i] != nums[i - 1])才进行递归。

public class Solution {
    /*
     * @param nums: A set of numbers.
     * @return: A list of lists. All valid subsets.
     */
    public List<List<Integer>> subsetsWithDup(int[] nums) {

        if (nums == null){
            return null;
        }

        //List<List<Integer>> result = new ArrayList<>(new ArrayList<Integer>());
        List<List<Integer>> result = new ArrayList<>();

        if (nums.length == 0){
            result.add(new ArrayList<Integer>());
            return result;
        }

        Arrays.sort(nums);
        helper(new ArrayList<Integer>(), nums, 0, result);
        return result;

    }

    private void helper(List<Integer> subset, int[] nums, int index, List<List<Integer>> result){

        result.add(new ArrayList<Integer>(subset));

        for (int i = index; i < nums.length; i++){
            if (i == index || nums[i] != nums[i - 1]){
                subset.add(nums[i]);
                helper(subset, nums, i + 1, result);
                subset.remove(subset.size() - 1);
            }
        }
    }
}