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, or perhaps an Integer[] or int[], with sequential values from some

Answer 1

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 2

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 3

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());

Answer 4

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 5

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 6

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());