How can I generate a list or array of sequential integers in Java?

Question

Is there a short and sweet way to generate a List<Integer>, or perhaps an Integer[] or int[], with sequential values from some start value to an end value?

That is, something shorter than, but equivalent to¹ the following:

void List<Integer> makeSequence(int begin, int end) {
  List<Integer> ret = new ArrayList<>(end - begin + 1);
  for (int i=begin; i<=end; i++) {
    ret.add(i);
  }
  return ret;  
}

The use of guava is fine.

Update:

Performance Analysis

Since this question has received several good answers, both using native Java 8 and third party libraries, I thought I'd test the performance of all the solutions.

The first test simply tests creating a list of 10 elements [1..10] using the following methods:

• classicArrayList: the code given above in my question (and essentially the same as adarshr's answer).
• eclipseCollections: the code given in Donald's answer below using Eclipse Collections 8.0.
• guavaRange: the code given in daveb's answer below. Technically, this doesn't create a List<Integer> but rather a ContiguousSet<Integer> - but since it implements Iterable<Integer> in-order, it mostly works for my purposes.
• intStreamRange: the code given in Vladimir's answer below, which uses IntStream.rangeClosed() - which was introduced in Java 8.
• streamIterate: the code given in Catalin's answer below which also uses IntStream functionality introduced in Java 8.

Here are the results in kilo-operations per second (higher numbers are better), for all the above with lists of size 10:

... and again for lists of size 10,000:

That last chart is correct - the solutions other than Eclipse and Guava are too slow to even get a single pixel bar! The fast solutions are 10,000 to 20,000 times faster than the rest.

What's going on here, of course, is that the guava and eclipse solutions don't actually materialize any kind of 10,000 element list - they are simply fixed-size wrappers around the start and endpoints. Each element is created as needed during iteration. Since we don't actually iterate in this test, the cost is deferred. All of the other solutions actually materialize the full list in memory and pay a heavy price in a creation-only benchmark.

Let's do something a bit more realistic and also iterate over all the integers, summing them. So in the case of the IntStream.rangeClosed variant, the benchmark looks like:

@Benchmark
public int intStreamRange() {
    List<Integer> ret = IntStream.rangeClosed(begin, end).boxed().collect(Collectors.toList());  

    int total = 0;
    for (int i : ret) {
        total += i;
    }
    return total;  
}

Here, the pictures changes a lot, although the non-materializing solutions are still the fastest. Here's length=10:

... and length = 10,000:

The long iteration over many elements evens things up a lot, but eclipse and guava remain more than twice as fast even on the 10,000 element test.

So if you really want a List<Integer>, eclipse collections seems like the best choice - but of course if you use streams in a more native way (e.g., forgetting .boxed() and doing a reduction in the primitive domain) you'll probably end up faster than all these variants.

---

¹ Perhaps with the exception of error handling, e.g., if end < begin, or if the size exceeds some implementation or JVM limits (e.g., arrays larger than 2^31-1.

Answer 1

With Java 8 it is so simple so it doesn't even need separate method anymore:

List<Integer> range = IntStream.rangeClosed(start, end)
    .boxed().collect(Collectors.toList());

And in Java 16 or later:

List<Integer> range = IntStream.rangeClosed(start, end)
    .boxed().toList();

Answer 2

int[] arr = IntStream.rangeClosed(2, 5).toArray();
System.out.println(Arrays.toString(arr));
// [2, 3, 4, 5]

Integer[] boxedArr = IntStream.rangeClosed(2, 5)
  .boxed().toArray(Integer[]::new);
System.out.println(Arrays.toString(boxedArr));

// Since Java 16
List<Integer> list1 = IntStream.rangeClosed(2, 5)
  .boxed().toList();
System.out.println(list1);

List<Integer> list2 = IntStream.rangeClosed(2, 5)
  .boxed().collect(Collectors.toList());
System.out.println(list2);

List<Integer> list3 = Arrays.asList(boxedArr);
System.out.println(list3);

List<Integer> list4 = new ArrayList<>();
IntStream.rangeClosed(2, 5).forEachOrdered(list4::add);
System.out.println(list4);

Answer 3

Well, this one liner might qualify (uses Guava Ranges)

ContiguousSet<Integer> integerList = ContiguousSet.create(Range.closedOpen(0, 10), DiscreteDomain.integers());
System.out.println(integerList);

This doesn't create a List<Integer>, but ContiguousSet offers much the same functionality, in particular implementing Iterable<Integer> which allows foreach implementation in the same way as List<Integer>.

In older versions (somewhere before Guava 14) you could use this:

ImmutableList<Integer> integerList = Ranges.closedOpen(0, 10).asSet(DiscreteDomains.integers()).asList();
System.out.println(integerList);

Both produce:

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Answer 4

You can use the Interval class from Eclipse Collections.

List<Integer> range = Interval.oneTo(10);
range.forEach(System.out::print);  // prints 12345678910

The Interval class is lazy, so doesn't store all of the values.

LazyIterable<Integer> range = Interval.oneTo(10);
System.out.println(range.makeString(",")); // prints 1,2,3,4,5,6,7,8,9,10

Your method would be able to be implemented as follows:

public List<Integer> makeSequence(int begin, int end) {
    return Interval.fromTo(begin, end);
}

If you would like to avoid boxing ints as Integers, but would still like a list structure as a result, then you can use IntList with IntInterval from Eclipse Collections.

public IntList makeSequence(int begin, int end) {
    return IntInterval.fromTo(begin, end);
}

IntList has the methods sum(), min(), minIfEmpty(), max(), maxIfEmpty(), average() and median() available on the interface.

Update for clarity: 11/27/2017

An Interval is a List<Integer>, but it is lazy and immutable. It is extremely useful for generating test data, especially if you deal a lot with collections. If you want you can easily copy an interval to a List, Set or Bag as follows:

Interval integers = Interval.oneTo(10);
Set<Integer> set = integers.toSet();
List<Integer> list = integers.toList();
Bag<Integer> bag = integers.toBag();

An IntInterval is an ImmutableIntList which extends IntList. It also has converter methods.

IntInterval ints = IntInterval.oneTo(10);
IntSet set = ints.toSet();
IntList list = ints.toList();
IntBag bag = ints.toBag();

An Interval and an IntInterval do not have the same equals contract.

Update for Eclipse Collections 9.0

You can now create primitive collections from primitive streams. There are withAll and ofAll methods depending on your preference. If you are curious, I explain why we have both here. These methods exist for mutable and immutable Int/Long/Double Lists, Sets, Bags and Stacks.

Assert.assertEquals(
        IntInterval.oneTo(10),
        IntLists.mutable.withAll(IntStream.rangeClosed(1, 10)));

Assert.assertEquals(
        IntInterval.oneTo(10),
        IntLists.immutable.withAll(IntStream.rangeClosed(1, 10)));

Note: I am a committer for Eclipse Collections

Answer 5

The following one-liner Java 8 version will generate [ 1, 2 ,3 ... 10 ]. The first arg of iterate is the first nr in the sequence, and the first arg of limit is the last number.

List<Integer> numbers = Stream.iterate(1, n -> n + 1)
                              .limit(10)
                              .collect(Collectors.toList());

Answer 6

This is the shortest I could get using Core Java.

List<Integer> makeSequence(int begin, int end) {
  List<Integer> ret = new ArrayList(end - begin + 1);

  for(int i = begin; i <= end; i++, ret.add(i));

  return ret;  
}

Answer 7

This is the shortest I could find.

List version

public List<Integer> makeSequence(int begin, int end)
{
    List<Integer> ret = new ArrayList<Integer>(++end - begin);

    for (; begin < end; )
        ret.add(begin++);

    return ret;
}

Array Version

public int[] makeSequence(int begin, int end)
{
    if(end < begin)
        return null;

    int[] ret = new int[++end - begin];
    for (int i=0; begin < end; )
        ret[i++] = begin++;
    return ret;
}

Answer 8

This one might works for you....

void List<Integer> makeSequence(int begin, int end) {

  AtomicInteger ai=new AtomicInteger(begin);
  List<Integer> ret = new ArrayList(end-begin+1);

  while ( end-->begin) {

    ret.add(ai.getAndIncrement());

  }
  return ret;  
}

Answer 9

You could use Guava Ranges

You can get a SortedSet by using

ImmutableSortedSet<Integer> set = Ranges.open(1, 5).asSet(DiscreteDomains.integers());
// set contains [2, 3, 4]