有没有办法在多个列表中找到共同的元素？

Question

I have a list of integer arrays. I need to find the common elements between those. What I can think of is an extension of what is listed in Common elements in two lists

Example would be 
[1,3,5],
[1,6,7,9,3],
[1,3,10,11]

should result in [1,3]

There are no duplicates in the arrays as well.

Is there a straight forward way to do this?

Answer 1

You can transform the lists to sets, and then use Set.retainAll method for intersection between the different sets. Once you intersect all sets, you are left with the common elements, and you can transform the resulting set back to a list.

Answer 2

You can use Set's intersection method offered by Guava , Here is a little example :

public <T> Set<T> intersection(List<T>... list) {
    Set<T> result = Sets.newHashSet(list[0]);
    for (List<T> numbers : list) {
        result = Sets.intersection(result, Sets.newHashSet(numbers));
    }
    return result;
}

Hope that could help you

Answer 3

We can use retainAll method of Collections . I initialised my commons arraylist with the first array list and called this for each remaining arraylists.

    List<List<Integer>> lists = new ArrayList<List<Integer>>();
    lists.add(new ArrayList<Integer>(Arrays.asList(1, 3, 5)));
    lists.add(new ArrayList<Integer>(Arrays.asList(1, 6, 7, 9, 3)));
    lists.add(new ArrayList<Integer>(Arrays.asList(1, 3, 10, 11)));

    List<Integer> commons = new ArrayList<Integer>();
    commons.addAll(lists.get(1));
    for (ListIterator<List<Integer>> iter = lists.listIterator(1); iter.hasNext(); ) {
        commons.retainAll(iter.next());
    }

    System.out.println(commons);
    System.out.println(lists.get(1));

Answer 4

with Java 8

ArrayList retain = list1.stream()
                     .filter(list2::contains).filter(list3::contains).collect(toList())

Answer 5

If you are looking for a function that returns elements that exist in all lists,

then the straight forward & simple way is building a statistic { < member, occurences > }

The condition here is no duplicates among the same list,

private Set<Integer> getCommonElements(ArrayList<Integer[]> idList)
{

    MapList<Integer,Short> stat = new MapList<Integer,Short>();

    // Here we count how many times each value occur
    for (int i = 0; i < idList.size(); i++)
    {
        for (int j = 0; j < idList.get(i).size; j++)
        {
            if (stat.containsKey(idList.get(i)[j]))
            {
                stat.set(idList.get(i)[j], stat.get(idList.get(i)[j])+1);
            }
            else
            {
                stat.add(idList.get(i)[j], 1);
            }
        }
    }

    // Here we only keep value that occured in all lists
    for (int i = 0; i < stat.size(); i++)
    {
        if (stat.get(i) < idList.size())
        {
            stat.remove(i);
            i--;
        }
    }

    return stat.keySet();
}

Answer 6

public class ArrayListImpl{
  public static void main(String s[]){
    ArrayList<Integer> al1=new ArrayList<Integer>();
     al1.add(21);al1.add(23);al1.add(25);al1.add(26);
    ArrayList<Integer> al2=new ArrayList<Integer>();
     al2.add(15);al2.add(16);al2.add(23);al2.add(25);
     ArrayList Al3=new ArrayList<Integer>();
     al3.addAll(al1);
      System.out.println("Al3 Elements :"+al3);
     al3.retainAll(al2); //Keeps common elements of (al1 & al2) & removes remaining elements
       System.out.println("Common Elements Between Two Array List:"+al3);  
}
}

Answer 7

public class commonvalue {

  
   Public static void MyMethod(){
        
       Set<integer> S1 = new set<integer>{1,3,5};
        Set<integer> S2 = new set<integer>{1,6,7,9,3};
        Set<integer> S3 = new set<integer>{1,3,10,11};
           
            s2.retainall(s1);
            s3.retainall(s2);
       
       system.debug(s3);
}
}

Answer 8

If you are using JAVA 8 streams. Then using stream reduce operation we can achieve the same.

Considering your example: Let's say a = [1,3,5], b = [1,6,7,9,3] and c = [1,3,10,11]

    List<Integer> commonElements = Stream.of(a,b,c)
         .reduce((s1,s2) -> {
              s1.retainAll(s2); 
              return s1;
          }).orElse(Collections.emptyList());

Keep in mind that after running this operation a will get modified with common values as well. So you will lose the actual value of a .

So elements of a and the result elements of commonElements will be essentially the same after running this operation.