打印列表的所有可能子集

Question

I have a List of elements (1, 2, 3), and I need to get the superset (powerset) of that list (without repeating elements). So basically I need to create a List of Lists that looks like:

{1}
{2}
{3}
{1, 2}
{1, 3}
{2, 3}
{1, 2, 3}

What is the best (simplicity > efficiency in this case, the list won't be huge) way to implement this? Preferably in Java, but a solution in any language would be useful.

Answer 1

Use bitmasks:

int allMasks = (1 << N);
for (int i = 1; i < allMasks; i++)
{
    for (int j = 0; j < N; j++)
        if ((i & (1 << j)) > 0) //The j-th element is used
           System.out.print((j + 1) + " ");

    System.out.println();
}

Here are all bitmasks:

1 = 001 = {1}
2 = 010 = {2}
3 = 011 = {1, 2}
4 = 100 = {3}
5 = 101 = {1, 3}
6 = 110 = {2, 3}
7 = 111 = {1, 2, 3}

You know in binary the first bit is the rightmost.

Answer 2

import java.io.*;
import java.util.*;
class subsets
{
    static String list[];
    public static void process(int n)
    {
        int i,j,k;
        String s="";
        displaySubset(s);
        for(i=0;i<n;i++)
        {
            for(j=0;j<n-i;j++)
            {
                k=j+i;
                for(int m=j;m<=k;m++)
                {
                    s=s+m;
                }
                displaySubset(s);
                s="";
            }
        }
    }
    public static void displaySubset(String s)
    {
        String set="";
        for(int i=0;i<s.length();i++)
        {
            String m=""+s.charAt(i);
            int num=Integer.parseInt(m);
            if(i==s.length()-1)
                set=set+list[num];
            else
                set=set+list[num]+",";
        }
        set="{"+set+"}";
        System.out.println(set);
    }
    public static void main()
    {
        Scanner sc=new Scanner(System.in);
        System.out.println("Input ur list");
        String slist=sc.nextLine();
        int len=slist.length();
        slist=slist.substring(1,len-1);
        StringTokenizer st=new StringTokenizer(slist,",");
        int n=st.countTokens();
        list=new String[n];
        for(int i=0;i<n;i++)
        {
            list[i]=st.nextToken();
        }
        process(n);
    }
}

Answer 3

A java solution based on Petar Minchev solution -

public static List<List<Integer>> getAllSubsets(List<Integer> input) {
    int allMasks = 1 << input.size();
    List<List<Integer>> output = new ArrayList<List<Integer>>();
    for(int i=0;i<allMasks;i++) {
        List<Integer> sub = new ArrayList<Integer>();
        for(int j=0;j<input.size();j++) {
            if((i & (1 << j)) > 0) {
                sub.add(input.get(j));
            }
        }
        output.add(sub);
    }

    return output;
}

Answer 4

I've noticed that answers are focused on the String list. Consequently, I decided to share more generic answer. Hope it'll be fouund helpful. (Soultion is based on another solutions I found, I combined it to a generic algorithem.)

/**
 * metod returns all the sublists of a given list
 * the method assumes all object are different
 * no matter the type of the list (generics)
 * @param list the list to return all the sublist of
 * @param <T>
 * @return list of the different sublists that can be made from the list object
 */
public static <T>  List<List<T>>getAllSubLists(List<T>list)
{
    List<T>subList;
    List<List<T>>res = new ArrayList<>();
    List<List<Integer>> indexes = allSubListIndexes(list.size());
    for(List<Integer> subListIndexes:indexes)
    {
        subList=new ArrayList<>();
        for(int index:subListIndexes)
            subList.add(list.get(index));
        res.add(subList);
    }
    return res;
}
/**
 * method returns list of list of integers representing the indexes of all the sublists in a N size list
 * @param n the size of the list
 * @return list of list of integers of indexes of the sublist
 */
public static List<List<Integer>> allSubListIndexes(int n) {
    List<List<Integer>> res = new ArrayList<>();
    int allMasks = (1 << n);
    for (int i = 1; i < allMasks; i++)
    {
        res.add(new ArrayList<>());
        for (int j = 0; j < n; j++)
            if ((i & (1 << j)) > 0)
                res.get(i-1).add(j);

    }
    return res;
}

Answer 5

This is the simple function can be used to create a list of all the possible numbers generated by digits of all possible subsets of the given array or list.

void SubsetNumbers(int[] arr){
    int len=arr.length;
    List<Integer> list=new ArrayList<Integer>();
    List<Integer> list1=new ArrayList<Integer>();
    for(int n:arr){
        if(list.size()!=0){
            for(int a:list){
                list1.add(a*10+n);
            }
            list1.add(n);
            list.addAll(list1);
            list1.clear();
        }else{
            list.add(n);
        }
    }
    System.out.println(list.toString());
}

Answer 6

In the given solution we iterate over every index and include current and all further elements.

class Solution {
        public List<List<Integer>> subsets(int[] nums) {
            List<List<Integer>> ans = new ArrayList<>();
            if(nums == null || nums.length ==0){
                return ans;
            }
            Arrays.sort(nums);
            List<Integer> subset = new ArrayList<>();
            allSubset(nums, ans , subset , 0);
            return ans;
        }
        private void allSubset(int[] nums,List<List<Integer>> ans ,List<Integer> subset , int idx){
            ans.add(new ArrayList<>(subset));
            for(int i = idx; i < nums.length; i++){
                subset.add(nums[i]);
                allSubset(nums, ans , subset , i+1);
                subset.remove(subset.size() - 1);
            }
        }
        
}

Answer 7

Peter Minchev's solution modified to handle larger lists through BigInteger

public static List<List<Integer>> getAllSubsets(List<Integer> input) {
    BigInteger allMasks = BigInteger.ONE.shiftLeft(input.size());
    List<List<Integer>> output = new ArrayList<>();
    for(BigInteger i=BigInteger.ZERO;allMasks.subtract(i).compareTo(BigInteger.ZERO)>0; i=i.add(BigInteger.ONE)) {
        List<Integer> subList = new ArrayList<Integer>();
        for(int j=0;j<input.size();j++) {
            if(i.and(BigInteger.valueOf(1<<j)).compareTo(BigInteger.ZERO) > 0) {
                subList.add(input.get(j));
            }
        }
        System.out.println(subList);
        output.add(subList);
    }
    return output;
}

Answer 8

/*---USING JAVA COLLECTIONS---*/
/*---O(n^3) Time complexity, Simple---*/

int[] arr = new int[]{1,2,3,4,5};
//Convert the array to ArrayList
List<Integer> arrList = new ArrayList<>();
for(int i=0;i<arr.length;i++)
    arrList.add(arr[i]);
List<List<Integer>> twoD_List = new ArrayList<>();
int k=1; /*-- k is used for toIndex in sublist() method---*/
while(k != arr.length+1) /*--- arr.length + 1 = toIndex for the last element---*/
    {
       for(int j=0;j<=arr.length-k;j++)
       {
          twoD_List.add(arrList.subList(j, j+k));/*--- fromIndex(j) - toIndex(j+k)...notice that j varies till (arr.length - k), while k is constant for the whole loop...k gets incremented after all the operations in this for loop---*/
       }
       k++; /*--- increment k for extending sublist(basically concept the toIndex)---*/
   }
//printing all sublists
for(List<Integer> list : twoD_List) System.out.println(list);