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Problem when working with arrays in Java - code does not go inside the "if" statement [duplicate]

How do I convert an array to a list in Java?
I used the Arrays.asList() but the behavior (and signature) somehow changed from Java SE 1.4.2 (docs now in archive) to 8 and most snippets I found on the web use the 1.4.2 behaviour.
For example:
int[] numbers = new int[] { 1, 2, 3 };
Arrays.asList(numbers)
on 1.4.2 returns a list containing the elements 1, 2, 3
on 1.5.0+ returns a list containing the array 'numbers'
In many cases it should be easy to detect, but sometimes it can slip unnoticed:
Assert.assertTrue(Arrays.asList(numbers).indexOf(4) == -1);

In your example, it is because you can't have a List of a primitive type. In other words, List<int> is not possible.
You can, however, have a List<Integer> using the Integer class that wraps the int primitive. Convert your array to a List with the Arrays.asList utility method.
Integer[] numbers = new Integer[] { 1, 2, 3 };
List<Integer> list = Arrays.asList(numbers);
See this code run live at IdeOne.com.

In Java 8, you can use streams:
int[] numbers = new int[] { 1, 2, 3 };
Arrays.stream(numbers)
.boxed()
.collect(Collectors.toList());

We cannot have List<int> as int is a primitive type so we can only have List<Integer>.
Java 16
Java 16 introduces a new method on Stream API called toList(). This handy method returns an unmodifiable List containing the stream elements. So, trying to add a new element to the list will simply lead to UnsupportedOperationException.
int[] ints = new int[] {1,2,3,4,5};
Arrays.stream(ints).boxed().toList();
Java 8 (int array)
int[] ints = new int[] {1,2,3,4,5};
List<Integer> list11 =Arrays.stream(ints).boxed().collect(Collectors.toList());
Java 8 and below (Integer array)
Integer[] integers = new Integer[] {1,2,3,4,5};
List<Integer> list21 = Arrays.asList(integers); // returns a fixed-size list backed by the specified array.
List<Integer> list22 = new ArrayList<>(Arrays.asList(integers)); // good
List<Integer> list23 = Arrays.stream(integers).collect(Collectors.toList()); //Java 8 only
Need ArrayList and not List?
In case we want a specific implementation of List e.g. ArrayList then we can use toCollection as:
ArrayList<Integer> list24 = Arrays.stream(integers)
.collect(Collectors.toCollection(ArrayList::new));
Why list21 cannot be structurally modified?
When we use Arrays.asList the size of the returned list is fixed because the list returned is not java.util.ArrayList, but a private static class defined inside java.util.Arrays. So if we add or remove elements from the returned list, an UnsupportedOperationException will be thrown. So we should go with list22 when we want to modify the list. If we have Java8 then we can also go with list23.
To be clear list21 can be modified in sense that we can call list21.set(index,element) but this list may not be structurally modified i.e. cannot add or remove elements from the list. You can also check this answer of mine for more explanation.
If we want an immutable list then we can wrap it as:
List<Integer> list22 = Collections.unmodifiableList(Arrays.asList(integers));
Another point to note is that the method Collections.unmodifiableList returns an unmodifiable view of the specified list. An unmodifiable view collection is a collection that is unmodifiable and is also a view onto a backing collection. Note that changes to the backing collection might still be possible, and if they occur, they are visible through the unmodifiable view.
We can have a truly immutable list in Java 9 and 10.
Truly Immutable list
Java 9:
String[] objects = {"Apple", "Ball", "Cat"};
List<String> objectList = List.of(objects);
Java 10 (Truly Immutable list):
We can use List.of introduced in Java 9. Also other ways:
List.copyOf(Arrays.asList(integers))
Arrays.stream(integers).collect(Collectors.toUnmodifiableList());

Speaking about conversion way, it depends on why do you need your List.
If you need it just to read data. OK, here you go:
Integer[] values = { 1, 3, 7 };
List<Integer> list = Arrays.asList(values);
But then if you do something like this:
list.add(1);
you get java.lang.UnsupportedOperationException.
So for some cases you even need this:
Integer[] values = { 1, 3, 7 };
List<Integer> list = new ArrayList<Integer>(Arrays.asList(values));
First approach actually does not convert array but 'represents' it like a List. But array is under the hood with all its properties like fixed number of elements. Please note you need to specify type when constructing ArrayList.

The problem is that varargs got introduced in Java 5 and unfortunately, Arrays.asList() got overloaded with a vararg version too. So Arrays.asList(numbers) is understood by the Java 5 compiler as a vararg parameter of int arrays.
This problem is explained in more details in Effective Java 2nd Ed., Chapter 7, Item 42.

I recently had to convert an array to a List. Later on the program filtered the list attempting to remove the data. When you use the Arrays.asList(array) function, you create a fixed size collection: you can neither add nor delete. This entry explains the problem better than I can: Why do I get an UnsupportedOperationException when trying to remove an element from a List?.
In the end, I had to do a "manual" conversion:
List<ListItem> items = new ArrayList<ListItem>();
for (ListItem item: itemsArray) {
items.add(item);
}
I suppose I could have added conversion from an array to a list using an List.addAll(items) operation.

Even shorter:
List<Integer> list = Arrays.asList(1, 2, 3, 4);

Using Arrays
This is the simplest way to convert an array to List. However, if you try to add a new element or remove an existing element from the list, an UnsupportedOperationException will be thrown.
Integer[] existingArray = {1, 2, 3};
List<Integer> list1 = Arrays.asList(existingArray);
List<Integer> list2 = Arrays.asList(1, 2, 3);
// WARNING:
list2.add(1); // Unsupported operation!
list2.remove(1); // Unsupported operation!
Using ArrayList or Other List Implementations
You can use a for loop to add all the elements of the array into a List implementation, e.g. ArrayList:
List<Integer> list = new ArrayList<>();
for (int i : new int[]{1, 2, 3}) {
list.add(i);
}
Using Stream API in Java 8
You can turn the array into a stream, then collect the stream using different collectors: The default collector in Java 8 use ArrayList behind the screen, but you can also impose your preferred implementation.
List<Integer> list1, list2, list3;
list1 = Stream.of(1, 2, 3).collect(Collectors.toList());
list2 = Stream.of(1, 2, 3).collect(Collectors.toCollection(ArrayList::new));
list3 = Stream.of(1, 2, 3).collect(Collectors.toCollection(LinkedList::new));
See also:
Why do we use autoboxing and unboxing in Java?
When to use LinkedList over ArrayList?

Another workaround if you use Apache commons-lang:
int[] numbers = new int[] { 1, 2, 3 };
Arrays.asList(ArrayUtils.toObject(numbers));
Where ArrayUtils.toObject converts int[] to Integer[]

In Java 9 you have the even more elegant solution of using immutable lists via the new convenience factory method List.of:
List<String> immutableList = List.of("one","two","three");
(shamelessly copied from here )

One-liner:
List<Integer> list = Arrays.asList(new Integer[] {1, 2, 3, 4});

If you are targeting Java 8 (or later), you can try this:
int[] numbers = new int[] {1, 2, 3, 4};
List<Integer> integers = Arrays.stream(numbers)
.boxed().collect(Collectors.<Integer>toList());
NOTE:
Pay attention to the Collectors.<Integer>toList(), this generic method helps you to avoid the error "Type mismatch: cannot convert from List<Object> to List<Integer>".

you have to cast in to array
Arrays.asList((Object[]) array)

Using Guava:
Integer[] array = { 1, 2, 3};
List<Integer> list = Lists.newArrayList(sourceArray);
Using Apache Commons Collections:
Integer[] array = { 1, 2, 3};
List<Integer> list = new ArrayList<>(6);
CollectionUtils.addAll(list, array);

I've had the same problem and wrote a generic function that takes an array and returns an ArrayList of the same type with the same contents:
public static <T> ArrayList<T> ArrayToArrayList(T[] array) {
ArrayList<T> list = new ArrayList<T>();
for(T elmt : array) list.add(elmt);
return list;
}

Given Array:
int[] givenArray = {2,2,3,3,4,5};
Converting integer array to Integer List
One way: boxed() -> returns the IntStream
List<Integer> givenIntArray1 = Arrays.stream(givenArray)
.boxed()
.collect(Collectors.toList());
Second Way: map each element of the stream to Integer and then collect
NOTE:
Using mapToObj you can covert each int element into string stream, char stream etc by casing i to (char)i
List<Integer> givenIntArray2 = Arrays.stream(givenArray)
.mapToObj(i->i)
.collect(Collectors.toList());
Converting One array Type to Another Type Example:
List<Character> givenIntArray2 = Arrays.stream(givenArray)
.mapToObj(i->(char)i)
.collect(Collectors.toList());

So it depends on which Java version you are trying-
Java 7
Arrays.asList(1, 2, 3);
OR
final String arr[] = new String[] { "G", "E", "E", "K" };
final List<String> initialList = new ArrayList<String>() {{
add("C");
add("O");
add("D");
add("I");
add("N");
}};
// Elements of the array are appended at the end
Collections.addAll(initialList, arr);
OR
Integer[] arr = new Integer[] { 1, 2, 3 };
Arrays.asList(arr);
In Java 8
int[] num = new int[] {1, 2, 3};
List<Integer> list = Arrays.stream(num)
.boxed().collect(Collectors.<Integer>toList())
Reference - http://www.codingeek.com/java/how-to-convert-array-to-list-in-java/

Can you improve this answer please as this is what I use but im not 100% clear. It works fine but intelliJ added new WeatherStation[0]. Why the 0 ?
public WeatherStation[] removeElementAtIndex(WeatherStation[] array, int index)
{
List<WeatherStation> list = new ArrayList<WeatherStation>(Arrays.asList(array));
list.remove(index);
return list.toArray(new WeatherStation[0]);
}

Use this to convert an Array arr to List.
Arrays.stream(arr).collect(Collectors.toList());
An example of defining a generic method to convert an array to a list:
public <T> List<T> fromArrayToList(T[] a) {
return Arrays.stream(a).collect(Collectors.toList());
}

use two line of code to convert array to list if you use it in integer value
you must use autoboxing type for primitive data type
Integer [] arr={1,2};
List<Integer> listInt=Arrays.asList(arr);

As of Java 8, the following should do
int[] temp = {1, 2, 3, 4, 5};
List<Integer> tempList = Arrays.stream(temp).boxed().collect(Collectors.toList());

If you are trying to optimize for memory, etc., (and don't want to pull in external libraries) it's simpler than you think to implement your own immutable "array view list" – you just need to extend java.util.AbstractList.
class IntArrayViewList extends AbstractList<Integer> {
int[] backingArray;
int size;
IntArrayViewList(int[] backingArray, int size) {
this.backingArray = backingArray;
this.size = size;
}
public Iterator<Integer> iterator() {
return new Iterator<Integer>() {
int i = 0;
#Override
public boolean hasNext() {
return i < size;
}
#Override
public Integer next() {
return get(i++);
}
};
}
public int size() {
return size;
}
public Integer get(int i) {
return backingArray[i];
}
}

int is a primitive. Primitives can’t accept null and have default value. Hence, to accept null you need to use wrapper class Integer.
Option 1:
int[] nos = { 1, 2, 3, 4, 5 };
Integer[] nosWrapped = Arrays.stream(nos).boxed()   
.toArray(Integer[]::new);
nosWrapped[5] = null // can store null
Option 2:
You can use any data structure that uses the wrapper class Integer
int[] nos = { 1, 2, 3, 4, 5 };
List<Integer> = Arrays.asList(nos)

I started looking at this by trying to reduce the amount of code preparing the input of some test cases. I see a lot of effort around trying to include advanced and new features along with Arrays.asList(), but below the code chosen due simplicity:
//Integer input[]
List<Integer> numbers = Arrays.asList(new Integer[]{1, 2 ,3, 4, 5, 4, 3, 2, 1, 3, 4});
//String input[]
List<String> names = Arrays.asList(new String[]{"Jhon", "Lucas", "Daniel", "Jim", "Sam"});
//String input[]
List<Character> letters = Arrays.asList(new Character[]{'A', 'B', 'K', 'J', 'F'});
Please notice that Anonymous array example will work just with Arrays of Non Primitive Types as the API uses Generics, that's the reason you can see several 2 line examples around, more info here: Why don't Java Generics support primitive types?
For newer JDKs there is another simpler option, the below examples are equivalent to the ones show above:
//Integer
List<Integer> numbers = Arrays.asList(1, 2 ,3, 4, 5, 4, 3, 2, 1, 3, 4);
//String
List<String> names = Arrays.asList("Jhon", "Lucas", "Daniel", "Jim", "Sam");
//Character
List<Character> letters = Arrays.asList('A', 'B', 'K', 'J', 'F');

Arrays.asList return type mismatch confusion [duplicate]

This question already has answers here:
Boxing with Arrays.asList()
(2 answers)
Closed 9 years ago.
Why does the following not return a list of integers?
int[] ints = new int[] { 1, 2, 3, 4, 5 };
List<Integer> intsList = Arrays.asList(ints); //compilation error
But instead a List of int[]
While this
String[] strings = new String[] { "Hello", "World" };
List<String> stringsList = Arrays.asList(strings);
Returns a list of String. I am guessing it fails due to it being an array of primitives but why? And how do I actually return a list of int.

It's because Arrays.asList(new int[] { 1, 2, 3, 4, 5 }) will create a List<int[]> with one item, not a List<Integer> with five items.
Note however that this would do what you expected:
List<Integer> intsList = Arrays.asList(1, 2, 3, 4, 5);
Your other alternatives are:
to create an Integer[] in the first place, or
to populate your list in a loop

The method is defined as:
public static <T> List<T> asList(T... a)
So in your first case, T is int[] and you are passing a single object to the method (i.e. the array), therefore it returns a list of int[].
I think you are mistaking with asList(1, 2, 3, 4, 5) (i.e. 5 items)

The T in List<T> must be some subtype of java.lang.Object, which int is not. THe only other interpretation is, as we are using ... that you are supplying an array of int[], i.e. an int[][]. So you get List<int[]>.
String is a subtype of Object, so this works as expected. Also the only way it can work prior to varargs introduction in J2SE 5.0. Generally the interpretation of existing code should not alter between language versions.
Now, if you wanted a List<Integer> you could go through and box each integer. If there a lot of these elements in your program then memory may be an issue. You may want to use a third-party library that compactly backs a List<Integer> with an int[], or just stick with arrays for primitives. It's unfortunate that Java does not support value types.

Remove duplicate in a list which contains arrays

I have a list<array[]> in Java, and I want to remove from it all duplicates.
In addition, the array [1,2] is the same as array [2,1].
I want to use Set, but as I understand it, if I declare 2 arrays: int[] array1 = {1, 2} and int[] array2 = {1,2}, Java considers them as 2 different arrays. In addition, It doesn't help me in the case of [1,2] and [2,1]
How can I do it?

I'm not tested but you can do sort the array before adding into set
int[] array1 = {1, 2};
Arrays.sort(array1);
set.add(array1);

Use a HashSet and then add all the contents of your current list(s) into it. Then use the .toArray() method to get back a list.
HashSet<Integer> noDuplicates = new HashSet<Integer>();
Integer[] array = { 1, 2, 2, 3, 3, 4, 1};
System.out.println(Arrays.toString(array));
for (Integer i : array)
noDuplicates.add(i);
array = noDuplicates.toArray(new Integer[] {});
System.out.println(Arrays.toString(array));
If you have to check if two arrays are the same often, then you might just want to use a Set instead of an array as it'll save you the extra step of putting everything into a Set and then checking.

Use Iterator to iterate the list.
Use Arrays.equals(arg0, arg1) method to compare arrays inside list.
If you find equal arrays remove using iterator.remove().

List<int[]> b = new ArrayList<int[]>(); // Your list here
for (int[] ar : b) {
Arrays.sort(ar);
}
Set<int[]> a = new HashSet<>(b);

Add multiple items to an already initialized arraylist in Java

My arraylist might be populated differently based on a user setting, so I've initialized it with
ArrayList<Integer> arList = new ArrayList<Integer>();
How can I add hundreds of integers without doing it one by one with arList.add(55);?

If you have another list that contains all the items you would like to add you can do arList.addAll(otherList). Alternatively, if you will always add the same elements to the list you could create a new list that is initialized to contain all your values and use the addAll() method, with something like
Integer[] otherList = new Integer[] {1, 2, 3, 4, 5};
arList.addAll(Arrays.asList(otherList));
or, if you don't want to create that unnecessary array:
arList.addAll(Arrays.asList(1, 2, 3, 4, 5));
Otherwise you will have to have some sort of loop that adds the values to the list individually.

What is the "source" of those integers? If it is something that you need to hard code in your source code, you may do
arList.addAll(Arrays.asList(1,1,2,3,5,8,13,21));

Collections.addAll is a varargs method which allows us to add any number of items to a collection in a single statement:
List<Integer> list = new ArrayList<>();
Collections.addAll(list, 1, 2, 3, 4, 5);
It can also be used to add array elements to a collection:
Integer[] arr = ...;
Collections.addAll(list, arr);

If you are looking to avoid multiple code lines to save space, maybe this syntax could be useful:
java.util.ArrayList lisFieldNames = new ArrayList() {
{
add("value1");
add("value2");
}
};
Removing new lines, you can show it compressed as:
java.util.ArrayList lisFieldNames = new ArrayList() {
{
add("value1"); add("value2"); (...);
}
};

Java 9+ now allows this:
List<Integer> arList = List.of(1,2,3,4,5);
The list will be immutable though.

In a Kotlin way;
val arList = ArrayList<String>()
arList.addAll(listOf(1,2,3,4,5))

If you needed to add a lot of integers it'd probably be easiest to use a for loop. For example, adding 28 days to a daysInFebruary array.
ArrayList<Integer> daysInFebruary = new ArrayList<>();
for(int i = 1; i <= 28; i++) {
daysInFebruary.add(i);
}

I believe scaevity's answer is incorrect. The proper way to initialize with multiple values would be this...
int[] otherList = {1,2,3,4,5};
So the full answer with the proper initialization would look like this
int[] otherList = {1,2,3,4,5};
arList.addAll(Arrays.asList(otherList));

How to convert an Array to a Set in Java

I would like to convert an array to a Set in Java. There are some obvious ways of doing this (i.e. with a loop) but I would like something a bit neater, something like:
java.util.Arrays.asList(Object[] a);
Any ideas?

Like this:
Set<T> mySet = new HashSet<>(Arrays.asList(someArray));
In Java 9+, if unmodifiable set is ok:
Set<T> mySet = Set.of(someArray);
In Java 10+, the generic type parameter can be inferred from the arrays component type:
var mySet = Set.of(someArray);
Be careful
Set.of throws IllegalArgumentException - if there are any duplicate
elements in someArray.
See more details: https://docs.oracle.com/en/java/javase/11/docs/api/java.base/java/util/Set.html#of(E...)

Set<T> mySet = new HashSet<T>();
Collections.addAll(mySet, myArray);
That's Collections.addAll(java.util.Collection, T...) from JDK 6.
Additionally: what if our array is full of primitives?
For JDK < 8, I would just write the obvious for loop to do the wrap and add-to-set in one pass.
For JDK >= 8, an attractive option is something like:
Arrays.stream(intArray).boxed().collect(Collectors.toSet());

With Guava you can do:
T[] array = ...
Set<T> set = Sets.newHashSet(array);

Java 8:
String[] strArray = {"eins", "zwei", "drei", "vier"};
Set<String> strSet = Arrays.stream(strArray).collect(Collectors.toSet());
System.out.println(strSet);
// [eins, vier, zwei, drei]

Varargs will work too!
Stream.of(T... values).collect(Collectors.toSet());

Java 8
We have the option of using Stream as well. We can get stream in various ways:
Set<String> set = Stream.of("A", "B", "C", "D").collect(Collectors.toCollection(HashSet::new));
System.out.println(set);
String[] stringArray = {"A", "B", "C", "D"};
Set<String> strSet1 = Arrays.stream(stringArray).collect(Collectors.toSet());
System.out.println(strSet1);
// if you need HashSet then use below option.
Set<String> strSet2 = Arrays.stream(stringArray).collect(Collectors.toCollection(HashSet::new));
System.out.println(strSet2);
The source code of Collectors.toSet() shows that elements are added one by one to a HashSet but specification does not guarantee it will be a HashSet.
"There are no guarantees on the type, mutability, serializability, or
thread-safety of the Set returned."
So it is better to use the later option. The output is:
[A, B, C, D]
[A, B, C, D]
[A, B, C, D]
Immutable Set (Java 9)
Java 9 introduced Set.of static factory method which returns immutable set for the provided elements or the array.
#SafeVarargs
static <E> Set<E> of​(E... elements)
Check Immutable Set Static Factory Methods for details.
Immutable Set (Java 10)
We can also get an immutable set in two ways:
Set.copyOf(Arrays.asList(array))
Arrays.stream(array).collect(Collectors.toUnmodifiableList());
The method Collectors.toUnmodifiableList() internally makes use of Set.of introduced in Java 9. Also check this answer of mine for more.

After you do Arrays.asList(array) you can execute Set set = new HashSet(list);
Here is a sample method, you can write:
public <T> Set<T> GetSetFromArray(T[] array) {
return new HashSet<T>(Arrays.asList(array));
}

There has been a lot of great answers already, but most of them won't work with array of primitives (like int[], long[], char[], byte[], etc.)
In Java 8 and above, you can box the array with:
Integer[] boxedArr = Arrays.stream(arr).boxed().toArray(Integer[]::new);
Then convert to set using stream:
Stream.of(boxedArr).collect(Collectors.toSet());

In Eclipse Collections, the following will work:
Set<Integer> set1 = Sets.mutable.of(1, 2, 3, 4, 5);
Set<Integer> set2 = Sets.mutable.of(new Integer[]{1, 2, 3, 4, 5});
MutableSet<Integer> mutableSet = Sets.mutable.of(1, 2, 3, 4, 5);
ImmutableSet<Integer> immutableSet = Sets.immutable.of(1, 2, 3, 4, 5);
Set<Integer> unmodifiableSet = Sets.mutable.of(1, 2, 3, 4, 5).asUnmodifiable();
Set<Integer> synchronizedSet = Sets.mutable.of(1, 2, 3, 4, 5).asSynchronized();
ImmutableSet<Integer> immutableSet = Sets.mutable.of(1, 2, 3, 4, 5).toImmutable();
Note: I am a committer for Eclipse Collections

Quickly : you can do :
// Fixed-size list
List list = Arrays.asList(array);
// Growable list
list = new LinkedList(Arrays.asList(array));
// Duplicate elements are discarded
Set set = new HashSet(Arrays.asList(array));
and to reverse
// Create an array containing the elements in a list
Object[] objectArray = list.toArray();
MyClass[] array = (MyClass[])list.toArray(new MyClass[list.size()]);
// Create an array containing the elements in a set
objectArray = set.toArray();
array = (MyClass[])set.toArray(new MyClass[set.size()]);

I've written the below from the advice above - steal it... it's nice!
/**
* Handy conversion to set
*/
public class SetUtil {
/**
* Convert some items to a set
* #param items items
* #param <T> works on any type
* #return a hash set of the input items
*/
public static <T> Set<T> asSet(T ... items) {
return Stream.of(items).collect(Collectors.toSet());
}
}

If you need to build an immutable set with only one element inside it, you can use Collections.singleton(...). Here is an example:
Set<String> mySet = Collections.singleton("Have a good day :-)");
This doesn't answer the original question but might be useful to someone (it would have been at least to me). If you think this answer does not fit just tell me and I will delete it.

Sometime using some standard libraries helps a lot. Try to look at the Apache Commons Collections. In this case your problems is simply transformed to something like this
String[] keys = {"blah", "blahblah"}
Set<String> myEmptySet = new HashSet<String>();
CollectionUtils.addAll(pythonKeywordSet, keys);
And here is the CollectionsUtils javadoc

Use CollectionUtils or ArrayUtils from stanford-postagger-3.0.jar
import static edu.stanford.nlp.util.ArrayUtils.asSet;
or
import static edu.stanford.nlp.util.CollectionUtils.asSet;
...
String [] array = {"1", "q"};
Set<String> trackIds = asSet(array);

In Java 10:
String[] strs = {"A", "B"};
Set<String> set = Set.copyOf(Arrays.asList(strs));
Set.copyOf returns an unmodifiable Set containing the elements of the given Collection.
The given Collection must not be null, and it must not contain any null elements.

private Map<Integer, Set<Integer>> nobreaks = new HashMap();
nobreaks.put(1, new HashSet(Arrays.asList(new int[]{2, 4, 5})));
System.out.println("expected size is 3: " +nobreaks.get(1).size());
the output is
expected size is 3: 1
change it to
nobreaks.put(1, new HashSet(Arrays.asList( 2, 4, 5 )));
the output is
expected size is 3: 3

For anyone solving for Android:
Kotlin Collections Solution
The asterisk * is the spread operator. It applies all elements in a collection individually, each passed in order to a vararg method parameter. It is equivalent to:
val myArray = arrayOf("data", "foo")
val mySet = setOf(*myArray)
// Equivalent to
val mySet = setOf("data", "foo")
// Multiple spreads ["data", "foo", "bar", "data", "foo"]
val mySet = setOf(*myArray, "bar", *myArray)
Passing no parameters setOf() results in an empty set.
In addition to setOf, you can also use any of these for a specific hash type:
hashSetOf()
linkedSetOf()
mutableSetOf()
sortableSetOf()
This is how to define the collection item type explicitly.
setOf<String>()
hashSetOf<MyClass>()

new HashSet<Object>(Arrays.asList(Object[] a));
But I think this would be more efficient:
final Set s = new HashSet<Object>();
for (Object o : a) { s.add(o); }

Set<T> b = new HashSet<>(Arrays.asList(requiredArray));