import java.util.ArrayList;import java.util.Arrays;import java.util.List;/** * Source : https://oj.leetcode.com/problems/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], *   [] * ] * */public class SubSet2 {    private List
    
     
      > result = null;    /**     *     *     * @return     */    public List
      
       
        > subset (int[] arr) {        result = new ArrayList
        
         
          >(); List
          
            set = new ArrayList
           
            (); Arrays.sort(arr); recursion(arr, 0, set); return result; } private void recursion (int[] arr, int index, List
            
              set) { if (index == arr.length) { return; } // 初始化上一次出栈的元素为当前将要进栈的元素-1，为了不和将要进栈的元素相同 int last = arr[index]-1; for (int i = index; i < arr.length; i++) { // 如果上一次出栈的元素等于准备进栈的元素，则说明两个元素一样，跳过 if (last == arr[i]) { continue; } set.add(arr[i]); List
             
               temp = new ArrayList
              
               (set); result.add(temp); recursion(arr, i + 1, set); last = set.remove(set.size()-1); } } private static void print (List
               
                
                 > list) { for (List
                 
                   arr : list) { System.out.println(Arrays.toString(arr.toArray(new Integer[arr.size()]))); } System.out.println(); } public static void main(String[] args) { SubSet2 subSet2 = new SubSet2(); int[] arr = new int[]{1,2,2}; int[] arr1 = new int[]{1,2,3,3,3,3,4}; print(subSet2.subset(arr)); print(subSet2.subset(arr1)); }}