Subsets II

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

Note:

• Elements in a subset must be in non-descending order.
• The solution set must not contain duplicate subsets.

 

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

[

  [2],

  [1],

  [1,2,2],

  [2,2],

  [1,2],

  []

]
和subsets 算法一致，不过在加入到result前 先判定是否在hashmap中。

 1 public class Solution {

 2     public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] S) {

 3         // IMPORTANT: Please reset any member data you declared, as

 4         // the same Solution instance will be reused for each test case.

 5         Arrays.sort(S);

 6         HashMap<ArrayList<Integer>, Integer> hm = new HashMap<ArrayList<Integer>, Integer>();

 7         ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();

 8         for(int i = 0; i < (1 << S.length); i ++){

 9             ArrayList<Integer> row = new ArrayList<Integer>();

10             for(int j = 0; j < S.length; j ++){

11                 if((i & (1 << j)) != 0) row.add(S[j]);

12             }

13             if(!hm.containsKey(row)){

14                 hm.put(row, 1);

15                 result.add(row);

16             }

17         }

18         return result;

19     }

20 }

 第三遍：

使用recursive 做：

 1 public class Solution {

 2     ArrayList<List<Integer>> result = null;

 3     public List<List<Integer>> subsetsWithDup(int[] num) {

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

 5         if(num == null || num.length == 0) return result;

 6         Arrays.sort(num);

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

 8         subset(num, 0, new ArrayList<Integer>());

 9         return result;

10     }

11     public void subset(int[] num, int pos, ArrayList<Integer> row){

12         int pre = Integer.MIN_VALUE;

13         for(int i = pos; i < num.length; i ++){

14             if(num[i] != pre){

15                 pre = num[i];

16                 ArrayList<Integer> nrow = new ArrayList<Integer> (row);

17                 nrow.add(pre);

18                 result.add(nrow);

19                 subset(num, i + 1, nrow);

20             }

21         }

22     }

23 }