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);
In Java 8:
List<String> newList = Stream.concat(listOne.stream(), listTwo.stream())
.collect(Collectors.toList());
Java 16+:
List<String> newList = Stream.concat(listOne.stream(), listTwo.stream()).toList();
Off the top of my head, I can shorten it by one line:
List<String> newList = new ArrayList<String>(listOne);
newList.addAll(listTwo);
addAll()
of both. I tried all those that suggest not copying the lists and they result having a lot of overhead we didn't need this time.
You could use the Apache commons-collections library:
List<String> newList = ListUtils.union(list1, list2);
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());
x -> x.stream()
could be replaced with Collection::stream
.
List::stream
.
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);
Collections.unmodifiableList()
method, which wraps a list to make it unmodifiable. CompositeUnmodifiableList
does the same thing, except it wraps two lists and provides a concatenated view. All the points you make about CompositeUnmodifiableList
are also true of Collections.unmodifiableList()
as well.
List<? extends E>
Probably not simpler, but intriguing and ugly:
List<String> newList = new ArrayList<String>() { { addAll(listOne); addAll(listTwo); } };
Don't use it in production code... ;)
Not simpler, but without resizing overhead:
List<String> newList = new ArrayList<>(listOne.size() + listTwo.size());
newList.addAll(listOne);
newList.addAll(listTwo);
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().
com.google.common.collect.Iterators#concat(java.util.Iterator<? extends java.util.Iterator<? extends T>>)
instead of Iterables#concat()
; because the later still copy elements into temp link!
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);
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.
newList
is not observable by any other thread. But you are correct that this probably shouldn't be done if it is unknown where the value of newList
came from (for example if newList
was passed as a parameter.
.forEach(newList::addAll);
instead of .collect(Collectors.toList());
?
List<List<Object>>
. What you may have in mind is something like this: stackoverflow.com/questions/189559/…
flatMap
.
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());
Slightly simpler:
List<String> newList = new ArrayList<String>(listOne);
newList.addAll(listTwo);
List
structure imposes no uniqueness constraints. You can remove dupes by doing the same thing with sets. Set<String> newSet = new HashSet<>(setOne); newSet.addAll(setTwo);
A little shorter would be:
List<String> newList = new ArrayList<String>(listOne);
newList.addAll(listTwo);
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());
}
You can do a oneliner if the target list is predeclared.
(newList = new ArrayList<String>(list1)).addAll(list2);
In Java 8 (the other way):
List<?> newList =
Stream.of(list1, list2).flatMap(List::stream).collect(Collectors.toList());
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]
flatMap
, because the lists are only iterated once when they are collected
flatMap
causes additional iteration, could you clarify?
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]
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
add
does not refer to the bulk addAll
operation.
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;
}
@SafeVarargs
!
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);
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());
}
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;
}
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)));
}
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;
}
My favourite way, using fluent api and Guava:
List<String> combined = ImmutableList.<String>builder().addAll(list1).addAll(list2).build()
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(),
...
}))
Vector
, take a 5 minute break.
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);
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());
}
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);
}}
);
Success story sharing
List<String> newList = Stream.concat(listOne.stream(), listTwo.stream()).distinct().collect(Collectors.toList());
Stream.of(listOne, listTwo).flatMap(Collection::stream).collect(Collectors.toList())