How do I join two lists in Java?

Question

Conditions: do not modify the original lists; JDK only, no external libraries. Bonus points for a one-liner or a JDK 1.3 version.

Is there a simpler way than:

List<String> newList = new ArrayList<String>();
newList.addAll(listOne);
newList.addAll(listTwo);

Answer 1

In Java 8:

List<String> newList = Stream.concat(listOne.stream(), listTwo.stream())
                             .collect(Collectors.toList());

Answer 2

Off the top of my head, I can shorten it by one line:

List<String> newList = new ArrayList<String>(listOne);
newList.addAll(listTwo);

Answer 3

You could use the Apache commons-collections library:

List<String> newList = ListUtils.union(list1, list2);

Answer 4

Another Java 8 one-liner:

List<String> newList = Stream.of(listOne, listTwo)
                            .flatMap(Collection::stream)
                            .collect(Collectors.toList());

As a bonus, since Stream.of() is variadic, you may concatenate as many lists as you like.

List<String> newList = Stream.of(listOne, listTwo, listThree)
                            .flatMap(Collection::stream)
                            .collect(Collectors.toList());

Answer 5

One of your requirements is to preserve the original lists. If you create a new list and use addAll(), you are effectively doubling the number of references to the objects in your lists. This could lead to memory problems if your lists are very large.

If you don't need to modify the concatenated result, you can avoid this using a custom list implementation. The custom implementation class is more than one line, obviously...but using it is short and sweet.

CompositeUnmodifiableList.java:

public class CompositeUnmodifiableList<E> extends AbstractList<E> {

    private final List<? extends E> list1;
    private final List<? extends E> list2;

    public CompositeUnmodifiableList(List<? extends E> list1, List<? extends E> list2) {
        this.list1 = list1;
        this.list2 = list2;
    }
    
    @Override
    public E get(int index) {
        if (index < list1.size()) {
            return list1.get(index);
        }
        return list2.get(index-list1.size());
    }

    @Override
    public int size() {
        return list1.size() + list2.size();
    }
}

Usage:

List<String> newList = new CompositeUnmodifiableList<String>(listOne,listTwo);

Answer 6

Probably not simpler, but intriguing and ugly:

List<String> newList = new ArrayList<String>() { { addAll(listOne); addAll(listTwo); } };

Don't use it in production code... ;)

Answer 7

Not simpler, but without resizing overhead:

List<String> newList = new ArrayList<>(listOne.size() + listTwo.size());
newList.addAll(listOne);
newList.addAll(listTwo);

Answer 8

Found this question looking to concatenate arbitrary amount of lists, not minding external libraries. So, perhaps it will help someone else:

com.google.common.collect.Iterables#concat()

Useful if you want to apply the same logic to a number of different collections in one for().

Answer 9

Java 8 (Stream.of and Stream.concat)

The proposed solution is for three lists though it can be applied for two lists as well. In Java 8 we can make use of Stream.of or Stream.concat as:

List<String> result1 = Stream.concat(Stream.concat(list1.stream(),list2.stream()),list3.stream()).collect(Collectors.toList());
List<String> result2 = Stream.of(list1,list2,list3).flatMap(Collection::stream).collect(Collectors.toList());

Stream.concat takes two streams as input and creates a lazily concatenated stream whose elements are all the elements of the first stream followed by all the elements of the second stream. As we have three lists we have used this method (Stream.concat) two times.

We can also write a utility class with a method that takes any number of lists (using varargs) and returns a concatenated list as:

public static <T> List<T> concatenateLists(List<T>... collections) {
        return Arrays.stream(collections).flatMap(Collection::stream).collect(Collectors.toList()); 
}

Then we can make use of this method as:

List<String> result3 = Utils.concatenateLists(list1,list2,list3);

Answer 10

Here is a java 8 solution using two lines:

List<Object> newList = new ArrayList<>();
Stream.of(list1, list2).forEach(newList::addAll);

Be aware that this method should not be used if

• the origin of newList is not known and it may already be shared with other threads
• the stream that modifies newList is a parallel stream and access to newList is not synchronized or threadsafe

due to side effect considerations.

Both of the above conditions do not apply for the above case of joining two lists, so this is safe.

Based on this answer to another question.

Answer 11

This is simple and just one line, but will add the contents of listTwo to listOne. Do you really need to put the contents in a third list?

Collections.addAll(listOne, listTwo.toArray());

Answer 12

Slightly simpler:

List<String> newList = new ArrayList<String>(listOne);
newList.addAll(listTwo);

Answer 13

A little shorter would be:

List<String> newList = new ArrayList<String>(listOne);
newList.addAll(listTwo);

Answer 14

You can create your generic Java 8 utility method to concat any number of lists.

@SafeVarargs
public static <T> List<T> concat(List<T>... lists) {
    return Stream.of(lists).flatMap(List::stream).collect(Collectors.toList());
}

Answer 15

You can do a oneliner if the target list is predeclared.

(newList = new ArrayList<String>(list1)).addAll(list2);

Answer 16

In Java 8 (the other way):

List<?> newList = 
Stream.of(list1, list2).flatMap(List::stream).collect(Collectors.toList());

Answer 17

another one liner solution using Java8 stream, since flatMap solution is already posted, here is a solution without flatMap

List<E> li = lol.stream().collect(ArrayList::new, List::addAll, List::addAll);

or

List<E> ints = Stream.of(list1, list2).collect(ArrayList::new, List::addAll, List::addAll);

code

    List<List<Integer>> lol = Arrays.asList(Arrays.asList(1, 2, 3), Arrays.asList(4, 5, 6));
    List<Integer> li = lol.stream().collect(ArrayList::new, List::addAll, List::addAll);
    System.out.println(lol);
    System.out.println(li);

output

[[1, 2, 3], [4, 5, 6]]
[1, 2, 3, 4, 5, 6]

Answer 18

We can join 2 lists using java8 with 2 approaches.

    List<String> list1 = Arrays.asList("S", "T");
    List<String> list2 = Arrays.asList("U", "V");

1) Using concat :

    List<String> collect2 = Stream.concat(list1.stream(), list2.stream()).collect(toList());
    System.out.println("collect2 = " + collect2); // collect2 = [S, T, U, V]

2) Using flatMap :

    List<String> collect3 = Stream.of(list1, list2).flatMap(Collection::stream).collect(toList());
    System.out.println("collect3 = " + collect3); // collect3 = [S, T, U, V]

Answer 19

Almost of answers suggest to use an ArrayList.

List<String> newList = new LinkedList<>(listOne);
newList.addAll(listTwo);

Prefer to use a LinkedList for efficient add operations.

ArrayList add is O(1) amortized, but O(n) worst-case since the array must be resized and copied. While LinkedList add is always constant O(1).

more infos https://stackoverflow.com/a/322742/311420

Answer 20

The smartest in my opinion:

/**
 * @param smallLists
 * @return one big list containing all elements of the small ones, in the same order.
 */
public static <E> List<E> concatenate (final List<E> ... smallLists)
{
    final ArrayList<E> bigList = new ArrayList<E>();
    for (final List<E> list: smallLists)
    {
        bigList.addAll(list);
    }
    return bigList;
}

Answer 21

You could do it with a static import and a helper class

nb the generification of this class could probably be improved

public class Lists {

   private Lists() { } // can't be instantiated

   public static List<T> join(List<T>... lists) {
      List<T> result = new ArrayList<T>();
      for(List<T> list : lists) {
         result.addAll(list);
      }
      return results;
   }

}

Then you can do things like

import static Lists.join;
List<T> result = join(list1, list2, list3, list4);

Answer 22

Java 8 version with support for joining by object key:

public List<SomeClass> mergeLists(final List<SomeClass> left, final List<SomeClass> right, String primaryKey) {
    final Map<Object, SomeClass> mergedList = new LinkedHashMap<>();

    Stream.concat(left.stream(), right.stream())
        .map(someObject -> new Pair<Object, SomeClass>(someObject.getSomeKey(), someObject))
        .forEach(pair-> mergedList.put(pair.getKey(), pair.getValue()));

    return new ArrayList<>(mergedList.values());
}

Answer 23

Use a Helper class.

I suggest:

public static <E> Collection<E> addAll(Collection<E> dest, Collection<? extends E>... src) {
    for(Collection<? extends E> c : src) {
        dest.addAll(c);
    }

    return dest;
}

public static void main(String[] args) {
    System.out.println(addAll(new ArrayList<Object>(), Arrays.asList(1,2,3), Arrays.asList("a", "b", "c")));

    // does not compile
    // System.out.println(addAll(new ArrayList<Integer>(), Arrays.asList(1,2,3), Arrays.asList("a", "b", "c")));

    System.out.println(addAll(new ArrayList<Integer>(), Arrays.asList(1,2,3), Arrays.asList(4, 5, 6)));
}

Answer 24

public static <T> List<T> merge(List<T>... args) {
    final List<T> result = new ArrayList<>();

    for (List<T> list : args) {
        result.addAll(list);
    }

    return result;
}

Answer 25

public static <T> List<T> merge(@Nonnull final List<T>... list) {
    // calculate length first
    int mergedLength = 0;
    for (List<T> ts : list) {
      mergedLength += ts.size();
    }

    final List<T> mergedList = new ArrayList<>(mergedLength);

    for (List<T> ts : list) {
      mergedList.addAll(ts);
    }

    return mergedList;
  }

Answer 26

I'm not claiming that it's simple, but you mentioned bonus for one-liners ;-)

Collection mergedList = Collections.list(new sun.misc.CompoundEnumeration(new Enumeration[] {
    new Vector(list1).elements(),
    new Vector(list2).elements(),
    ...
}))

Answer 27

Here's an approach using streams and java 8 if your lists have different types and you want to combine them to a list of another type.

public static void main(String[] args) {
    List<String> list2 = new ArrayList<>();
    List<Pair<Integer, String>> list1 = new ArrayList<>();

    list2.add("asd");
    list2.add("asdaf");
    list1.add(new Pair<>(1, "werwe"));
    list1.add(new Pair<>(2, "tyutyu"));

    Stream stream = Stream.concat(list1.stream(), list2.stream());

    List<Pair<Integer, String>> res = (List<Pair<Integer, String>>) stream
            .map(item -> {
                if (item instanceof String) {
                    return new Pair<>(0, item);
                }
                else {
                    return new Pair<>(((Pair<Integer, String>)item).getKey(), ((Pair<Integer, String>)item).getValue());
                }
            })
            .collect(Collectors.toList());
}

Answer 28

If you want to do this statically you can the following.

The examples uses 2 EnumSets in natural-order (==Enum-order) A, B and joins then in an ALL list.

public static final EnumSet<MyType> CATEGORY_A = EnumSet.of(A_1, A_2);
public static final EnumSet<MyType> CATEGORY_B = EnumSet.of(B_1, B_2, B_3);

public static final List<MyType> ALL = 
              Collections.unmodifiableList(
                  new ArrayList<MyType>(CATEGORY_A.size() + CATEGORY_B.size())
                  {{
                      addAll(CATEGORY_A);
                      addAll(CATEGORY_B);
                  }}
              );

Answer 29

No way near one-liner, but I think this is the simplest:

List<String> newList = new ArrayList<String>(l1);
newList.addAll(l2);

for(String w:newList)
        System.out.printf("%s ", w);

Answer 30

import java.util.AbstractList;
import java.util.List;


/**
 * The {@code ConcatList} is a lightweight view of two {@code List}s.
 * <p>
 * This implementation is <em>not</em> thread-safe even though the underlying lists can be.
 * 
 * @param <E>
 *            the type of elements in this list
 */
public class ConcatList<E> extends AbstractList<E> {

    /** The first underlying list. */
    private final List<E> list1;
    /** The second underlying list. */
    private final List<E> list2;

    /**
     * Constructs a new {@code ConcatList} from the given two lists.
     * 
     * @param list1
     *            the first list
     * @param list2
     *            the second list
     */
    public ConcatList(final List<E> list1, final List<E> list2) {
        this.list1 = list1;
        this.list2 = list2;
    }

    @Override
    public E get(final int index) {
        return getList(index).get(getListIndex(index));
    }

    @Override
    public E set(final int index, final E element) {
        return getList(index).set(getListIndex(index), element);
    }

    @Override
    public void add(final int index, final E element) {
        getList(index).add(getListIndex(index), element);
    }

    @Override
    public E remove(final int index) {
        return getList(index).remove(getListIndex(index));
    }

    @Override
    public int size() {
        return list1.size() + list2.size();
    }

    @Override
    public boolean contains(final Object o) {
        return list1.contains(o) || list2.contains(o);
    }

    @Override
    public void clear() {
        list1.clear();
        list2.clear();
    }

    /**
     * Returns the index within the corresponding list related to the given index.
     * 
     * @param index
     *            the index in this list
     * 
     * @return the index of the underlying list
     */
    private int getListIndex(final int index) {
        final int size1 = list1.size();
        return index >= size1 ? index - size1 : index;
    }

    /**
     * Returns the list that corresponds to the given index.
     * 
     * @param index
     *            the index in this list
     * 
     * @return the underlying list that corresponds to that index
     */
    private List<E> getList(final int index) {
        return index >= list1.size() ? list2 : list1;
    }

}

Answer 31

I can't improve on the two-liner in the general case without introducing your own utility method, but if you do have lists of Strings and you're willing to assume those Strings don't contain commas, you can pull this long one-liner:

List<String> newList = new ArrayList<String>(Arrays.asList((listOne.toString().subString(1, listOne.length() - 1) + ", " + listTwo.toString().subString(1, listTwo.length() - 1)).split(", ")));

If you drop the generics, this should be JDK 1.4 compliant (though I haven't tested that). Also not recommended for production code ;-)

Answer 32

public class TestApp {

/**
 * @param args
 */
public static void main(String[] args) {
    System.out.println("Hi");
    Set<List<String>> bcOwnersList = new HashSet<List<String>>();
    List<String> bclist = new ArrayList<String>();
    List<String> bclist1 = new ArrayList<String>();
    List<String> object = new ArrayList<String>();
    object.add("BC11");
    object.add("C2");
    bclist.add("BC1");
    bclist.add("BC2");
    bclist.add("BC3");
    bclist.add("BC4");
    bclist.add("BC5");
    bcOwnersList.add(bclist);
    bcOwnersList.add(object);

    bclist1.add("BC11");
    bclist1.add("BC21");
    bclist1.add("BC31");
    bclist1.add("BC4");
    bclist1.add("BC5");

    List<String> listList= new ArrayList<String>();
    for(List<String> ll : bcOwnersList){
        listList = (List<String>) CollectionUtils.union(listList,CollectionUtils.intersection(ll, bclist1));
    }
    /*for(List<String> lists : listList){
        test = (List<String>) CollectionUtils.union(test, listList);
    }*/
    for(Object l : listList){
        System.out.println(l.toString());
    }
    System.out.println(bclist.contains("BC"));

}

}