How to make a new List in Java

Question

We create a Set as:

Set myset = new HashSet()

How do we create a List in Java?

Answer 1

List myList = new ArrayList();

or with generics (Java 7 or later)

List<MyType> myList = new ArrayList<>();

or with generics (Old java versions)

List<MyType> myList = new ArrayList<MyType>();

Answer 2

Additionally, if you want to create a list that has things in it (though it will be fixed size):

List<String> messages = Arrays.asList("Hello", "World!", "How", "Are", "You");

Answer 3

Let me summarize and add something:

JDK

1. new ArrayList<String>();
2. Arrays.asList("A", "B", "C")

Guava

1. Lists.newArrayList("Mike", "John", "Lesly");
2. Lists.asList("A","B", new String [] {"C", "D"});

Immutable List

1. Collections.unmodifiableList(new ArrayList<String>(Arrays.asList("A","B")));
2. ImmutableList.builder()                                      // Guava
            .add("A")
            .add("B").build();
3. ImmutableList.of("A", "B");                                  // Guava
4. ImmutableList.copyOf(Lists.newArrayList("A", "B", "C"));     // Guava

Empty immutable List

1. Collections.emptyList();
2. Collections.EMPTY_LIST;

List of Characters

1. Lists.charactersOf("String")                                 // Guava
2. Lists.newArrayList(Splitter.fixedLength(1).split("String"))  // Guava

List of Integers

Ints.asList(1,2,3);                                             // Guava

Answer 4

In Java 8

To create a non-empty list of fixed size (operations like add, remove, etc., are not supported):

List<Integer> list = Arrays.asList(1, 2); // but, list.set(...) is supported

To create a non-empty mutable list:

List<Integer> list = new ArrayList<>(Arrays.asList(3, 4));

In Java 9

Using a new List.of(...) static factory methods:

List<Integer> immutableList = List.of(1, 2);

List<Integer> mutableList = new ArrayList<>(List.of(3, 4));

In Java 10

Using the Local Variable Type Inference:

var list1 = List.of(1, 2);

var list2 = new ArrayList<>(List.of(3, 4));

var list3 = new ArrayList<String>();

And follow best practices...

Don't use raw types

Since Java 5, generics have been a part of the language - you should use them:

List<String> list = new ArrayList<>(); // Good, List of String

List list = new ArrayList(); // Bad, don't do that!

Program to interfaces

For example, program to the List interface:

List<Double> list = new ArrayList<>();

Instead of:

ArrayList<Double> list = new ArrayList<>(); // This is a bad idea!

Answer 5

First read this, then read this and this. 9 times out of 10 you'll use one of those two implementations.

In fact, just read Sun's Guide to the Collections framework.

Answer 6

//simple example creating a list form a string array

String[] myStrings = new String[] {"Elem1","Elem2","Elem3","Elem4","Elem5"};

List mylist = Arrays.asList(myStrings );

//getting an iterator object to browse list items

Iterator itr= mylist.iterator();

System.out.println("Displaying List Elements,");

while(itr.hasNext())

  System.out.println(itr.next());

Answer 7

Since Java 7 you have type inference for generic instance creation, so there is no need to duplicate generic parameters on the right hand side of the assignment:

List<String> list = new ArrayList<>();

A fixed-size list can be defined as:

List<String> list = Arrays.asList("foo", "bar");

For immutable lists you can use the Guava library:

List<String> list = ImmutableList.of("foo", "bar");

Answer 8

List is just an interface just as Set.

Like HashSet is an implementation of a Set which has certain properties in regards to add / lookup / remove performance, ArrayList is the bare implementation of a List.

If you have a look at the documentation for the respective interfaces you will find "All Known Implementing Classes" and you can decide which one is more suitable for your needs.

Chances are that it's ArrayList.

Answer 9

List is an interface like Set and has ArrayList and LinkedList as general purpose implementations.

We can create List as:

 List<String> arrayList = new ArrayList<>();
 List<String> linkedList = new LinkedList<>(); 

We can also create a fixed-size list as:

List<String> list = Arrays.asList("A", "B", "C");

We would almost always be using ArrayList opposed to LinkedList implementation:

1. LinkedList uses a lot of space for objects and performs badly when we have lots of elements.
2. Any indexed operation in LinkedList requires O(n) time compared to O(1) in ArrayList.
3. Check this link for more information.

The list created by Arrays.asList above can not be modified structurally but its elements can still be modified.

Java 8

As per doc, the method Collections.unmodifiableList returns an unmodifiable view of the specified list. We can get it like:

Collections.unmodifiableList(Arrays.asList("A", "B", "C"));

Java 9

In case we are using Java 9 then:

List<String> list = List.of("A", "B");

Java 10

In case we are at Java 10 then the method Collectors.unmodifiableList will return an instance of truly unmodifiable list introduced in Java 9. Check this answer for more info about the difference in Collections.unmodifiableList vs Collectors.unmodifiableList in Java 10.

Answer 10

Sometimes - but only very rarely - instead of a new ArrayList, you may want a new LinkedList. Start out with ArrayList and if you have performance problems and evidence that the list is the problem, and a lot of adding and deleting to that list - then - not before - switch to a LinkedList and see if things improve. But in the main, stick with ArrayList and all will be fine.

Answer 11

List list = new ArrayList();

Or with generics

List<String> list = new ArrayList<String>();

You can, of course, replace string with any type of variable, such as Integer, also.

Answer 12

One example:

List somelist = new ArrayList();

You can look at the javadoc for List and find all known implementing classes of the List interface that are included with the java api.

Answer 13

Using Google Collections, you could use the following methods in the Lists class

import com.google.common.collect.Lists;

// ...

List<String> strings = Lists.newArrayList();

List<Integer> integers = Lists.newLinkedList();

There are overloads for varargs initialization and initialising from an Iterable<T>.

The advantage of these methods is that you don't need to specify the generic parameter explicitly as you would with the constructor - the compiler will infer it from the type of the variable.

Answer 14

As an option you can use double brace initialization here:

List<String> list = new ArrayList<String>(){
  {
   add("a");
   add("b");
  }
};

Answer 15

As declaration of array list in java is like

public class ArrayList<E> extends AbstractList<E> implements List<E>, RandomAccess, Cloneable, Serializable  

There is numerous way you can create and initialize array list in java.

 1) List list = new ArrayList();

 2) List<type> myList = new ArrayList<>();

 3) List<type> myList = new ArrayList<type>();

 4) Using Utility class

    List<Integer> list = Arrays.asList(8, 4);
    Collections.unmodifiableList(Arrays.asList("a", "b", "c"));

 5) Using static factory method

    List<Integer> immutableList = List.of(1, 2);


 6) Creation and initializing at a time

    List<String> fixedSizeList = Arrays.asList(new String[] {"Male", "Female"});



 Again you can create different types of list. All has their own characteristics

 List a = new ArrayList();
 List b = new LinkedList();
 List c = new Vector(); 
 List d = new Stack(); 
 List e = new CopyOnWriteArrayList();

Answer 16

List<Object> nameOfList = new ArrayList<Object>();

You need to import List and ArrayList.

Answer 17

More options to do the same thing with Java 8, not better, not worse, just different and if you want to do some extra work with the lists, Streams will provide you more alternatives (filter, map, reduce, etc.)

List<String> listA = Stream.of("a", "B", "C").collect(Collectors.toList());
List<Integer> listB = IntStream.range(10, 20).boxed().collect(Collectors.toList());
List<Double> listC = DoubleStream.generate(() -> { return new Random().nextDouble(); }).limit(10).boxed().collect(Collectors.toList());
LinkedList<Integer> listD = Stream.iterate(0, x -> x++).limit(10).collect(Collectors.toCollection(LinkedList::new));

Answer 18

With Java 9, you are able to do the following to create an immutable List:

List<Integer> immutableList = List.of(1, 2, 3, 4, 5);

List<Integer> mutableList = new ArrayList<>(immutableList);

Answer 19

There are many ways to create a Set and a List. HashSet and ArrayList are just two examples. It is also fairly common to use generics with collections these days. I suggest you have a look at what they are

This is a good introduction for java's builtin collections. http://java.sun.com/javase/6/docs/technotes/guides/collections/overview.html

Answer 20

List arrList = new ArrayList();

Its better you use generics as suggested below:

List<String> arrList = new ArrayList<String>();

arrList.add("one");

Incase you use LinkedList.

List<String> lnkList = new LinkedList<String>();

Answer 21

The following are some ways you can create lists.

• This will create a list with fixed size, adding/removing elements is not possible, it will throw a java.lang.UnsupportedOperationException if you try to do so.

  List<String> fixedSizeList = Arrays.asList(new String[] {"Male", "Female"});
  

• The following version is a simple list where you can add/remove any number of elements.

  List<String> list = new ArrayList<>();
  

• This is how to create a LinkedList in java, If you need to do frequent insertion/deletion of elements on the list, you should use LinkedList instead of ArrayList

  List<String> linkedList = new LinkedList<>();
  

Answer 22

Using Eclipse Collections you can create a List like this:

List<String> list1 = Lists.mutable.empty();
List<String> list2 = Lists.mutable.of("One", "Two", "Three");

If you want an immutable list:

ImmutableList<String> list3 = Lists.immutable.empty();
ImmutableList<String> list4 = Lists.immutable.of("One", "Two", "Three");

You can avoid auto-boxing by using primitive lists. Here's how you'd create int lists:

MutableIntList list5 = IntLists.mutable.empty();
MutableIntList list6 = IntLists.mutable.of(1, 2, 3);

ImmutableIntList list7 = IntLists.immutable.empty();
ImmutableIntList list8 = IntLists.immutable.of(1, 2, 3);

There are variants for all 8 primitives.

MutableLongList longList       = LongLists.mutable.of(1L, 2L, 3L);
MutableCharList charList       = CharLists.mutable.of('a', 'b', 'c');
MutableShortList shortList     = ShortLists.mutable.of((short) 1, (short) 2, (short) 3);
MutableByteList byteList       = ByteLists.mutable.of((byte) 1, (byte) 2, (byte) 3);
MutableBooleanList booleanList = BooleanLists.mutable.of(true, false);
MutableFloatList floatList     = FloatLists.mutable.of(1.0f, 2.0f, 3.0f);
MutableDoubleList doubleList   = DoubleLists.mutable.of(1.0, 2.0, 3.0);

Note: I am a committer for Eclipse Collections.

Answer 23

Try this:

List<String> messages = Arrays.asList("bla1", "bla2", "bla3");

Or:

List<String> list1 = Lists.mutable.empty(); // Empty
List<String> list2 = Lists.mutable.of("One", "Two", "Three");

Answer 24

If you need a serializable, immutable list with a single entity you can use:

List<String> singList = Collections.singletonList("stackoverlow");