Lintcode 18 Subsets II__Lintcode

Question

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

Recursion: 與Subsets 17類似，但是因為有重複元素，為了避免出現重複解，有幾點需要注意。

1) 將所有重複元素排在一起，因此首先對輸入數組進行排序

2) 避免重複的方法為：重複元素只能取第一個元素，因此在加入元素前要判斷該元素值是否和之前元素相等。

Non-Recursion:

代碼如下：

Recursion:

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