题目：

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：

在subsets的代码上加一句话if (i > start && nums[i] == nums[i - 1]) continue;

package recursion;import java.util.ArrayList;import java.util.Arrays;import java.util.List;public class SubsetsII {    public List
     
      
       > subsetsWithDup(int[] nums) {        List
       
        
         > res = new ArrayList
         
          
           >(); List
           
             record = new ArrayList
            
             (); Arrays.sort(nums); int n = nums.length; for (int k = 0; k <= n; ++k) generateRecord(res, record, nums, 0, n - 1, k); return res; } private void generateRecord(List
             
              
               > res, List
               
                 record, int[] nums, int start, int end, int k) { if (k == 0) { res.add(record); return; } for (int i = start; i <= end - k + 1; ++i) { if (i > start && nums[i] == nums[i - 1]) continue; List
                
                  newRecord = new ArrayList
                 
                  (record); newRecord.add(nums[i]); generateRecord(res, newRecord, nums, i + 1, end, k - 1); } } public static void main(String[] args) { // TODO Auto-generated method stub SubsetsII s = new SubsetsII(); int[] nums = { 1, 2, 2, 3 }; List
                  
                   
                    > res = s.subsetsWithDup(nums); for (List
                    
                      l : res) { for (int i : l) System.out.print(i + "\t"); System.out.println(); } }}