leetcode--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],
  []
]

A shorter solution:

public class Solution {
    public List<List<Integer>> subsetsWithDup(int[] num) {
        Arrays.sort(num);
		int length = num.length;
		List<List<Integer> > result = new ArrayList<List<Integer>>();
		List<Integer> aSubset = new ArrayList<Integer>();
		result.add(aSubset);
		int size = 1;
		int i = 0;
		while(i < length){
			int mult = 0;
			int current = num[i];
			while(i < length){
				if(current != num[i])
					break;
				else{
					++i;
					++mult;
				}
			}
			for(int k = 0; k < mult; ++k){
				for(int j = 0; j < size; ++j){
					List<Integer> newList = new ArrayList<Integer>();
					newList.addAll(result.get(k*size + j));
					newList.add(current);
					result.add(newList);
				}
			}
			size *= (mult + 1);
		}
		return result;     
    }
}

　

Another solution:　

public class Solution {
    public ArrayList<ArrayList<Integer> > subsetsWithDup(int[] num){
		Arrays.sort(num);
		HashMap<Integer, Integer> mult = new HashMap<Integer, Integer>();
		HashMap<Integer, Integer> dist = new HashMap<Integer, Integer>();
		int l = 0;
		for(int i = 0; i < num.length; ++i){
			if(mult.containsKey(num[i])){
				int mu = mult.get(num[i]);
				mult.put(num[i], ++mu);
			}
			else{
				mult.put(num[i], 1);
				dist.put(l, num[i]);
				++l;
			}
		}
		ArrayList<ArrayList<Integer> > result = new ArrayList<ArrayList<Integer> >();
		int nums = (int)Math.pow(2, l);
		for(int i = 0; i < nums; ++i){
			ArrayList<ArrayList<Integer> > list = new ArrayList<ArrayList<Integer> >();
			list.add(new ArrayList<Integer>());
			String binaryreps = Integer.toBinaryString(i);
			int length = binaryreps.length(); 
			for(int j = 0; j < length; ++j){
				ArrayList<ArrayList<Integer> > temp = new ArrayList<ArrayList<Integer> >();
				if(binaryreps.charAt(length - 1 - j) == '1'){
					while(!list.isEmpty()){
						ArrayList<Integer> aList = list.remove(0);
						int times = mult.get(dist.get(j));
						for(int m = 1; m <= times; ++m){
							ArrayList<Integer> newList = new ArrayList<Integer>();
							newList.addAll(aList);
							for(int n = 1; n <= m; ++n)
								newList.add(dist.get(j));
							temp.add(newList);
						}						
					}
					list = temp;
				}
			}
			result.addAll(list);	
		}		
		return result;
    }
}