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。

public class Solution {
    public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] S) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        Arrays.sort(S);
        HashMap<ArrayList<Integer>, Integer> hm = new HashMap<ArrayList<Integer>, Integer>();
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        for(int i = 0; i < (1 << S.length); i ++){
            ArrayList<Integer> row = new ArrayList<Integer>();
            for(int j = 0; j < S.length; j ++){
                if((i & (1 << j)) != 0) row.add(S[j]);
            }
            if(!hm.containsKey(row)){
                hm.put(row, 1);
                result.add(row);
            }
        }
        return result;
    }
}

 第三遍：

使用recursive 做：

public class Solution {
    ArrayList<List<Integer>> result = null;
    public List<List<Integer>> subsetsWithDup(int[] num) {
        result = new ArrayList<List<Integer>> ();
        if(num == null || num.length == 0) return result;
        Arrays.sort(num);
        result.add(new ArrayList<Integer>());
        subset(num, 0, new ArrayList<Integer>());
        return result;
    }
    public void subset(int[] num, int pos, ArrayList<Integer> row){
        int pre = Integer.MIN_VALUE;
        for(int i = pos; i < num.length; i ++){
            if(num[i] != pre){
                pre = num[i];
                ArrayList<Integer> nrow = new ArrayList<Integer> (row);
                nrow.add(pre);
                result.add(nrow);
                subset(num, i + 1, nrow);
            }
        }
    }
}