Given a collection of integers that might contain duplicates, nums, 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 nums = [1,2,2],
a solution is:

[

  [2],

  [1],

  [1,2,2],

  [2,2],

  [1,2],

  []

]

思路：这一题比subsets多一了一道反复，详细代码例如以下：

public class Solution {

    boolean[] b;

    Set<String> set;

    List<List<Integer>> list;

    Set<String> set1;

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

        b = new boolean[nums.length];

        set = new HashSet<String>();

        list = new ArrayList<List<Integer>>();

        set1 = new HashSet<String>();

        Arrays.sort(nums);

        count(nums,"",nums.length,0);

        return list;

    }

    private void count(int[] nums,String s,int n,int j){

            //没有反复才加入

            if(set.add(s)){

               //以","切割数组

               String[] sa = s.split(",");

               List<Integer> al = new ArrayList<Integer>();

               for(int i = 0; i < sa.length; i++){

            	   if(sa[i].length() > 0){

            		   al.add(Integer.parseInt(sa[i]));

            	   }

               }

               Collections.sort(al);

               if(set1.add(al.toString()))

            	   list.add(al);

            }

        for(int i = j; i < nums.length;i++){

            if(!b[i]){

                b[i] = true;

                count(nums,s + "," + nums[i],n-1,i+1);

                b[i] = false;

            }

        }

    }

}

以下这样的写法更简洁：

public class Solution {

    List<List<Integer>> list;//结果集

    List<Integer> al;//每一项

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

        list = new ArrayList<List<Integer>>();

        al = new ArrayList<Integer>();

        Arrays.sort(nums);

        //排列组合

        count(nums,al,0);

        return list;

    }

    private void count(int[] nums,List<Integer> al,int j){

    	list.add(new ArrayList<Integer>(al));//不反复的才加入

        for(int i = j; i < nums.length;i++){

        	if(i == j || nums[i] != nums[i-1]){//去除反复

        		al.add(nums[i]);//加入

                count(nums,al,i+1);

                al.remove(al.size()-1);//去除。为下一个结果做准备

        	}

        }

    }

}