Java: Interleaving multiple arrays into a single array

For simplicity, assume that the arrays are the same length, and are int arrays.

int[] merge(int[] a, int[] b)
{
    assert (a.length == b.length);

    int[] result = new int[a.length + b.length];

    for (int i=0; i<a.length; i++)
    {
        result[i*2] = a[i];
        result[i*2+1] = b[i];
    }

    return result;
}

I think this is not doable with your given constraints (O(n) time and O(1) space, i.e. no additional space) for an array or array-based list. (Assuming of course, that we can't simply create a new List object delegating to the original ones.)

If you have two linked lists, this is doable - if we assume the garbage collector is fast enough, i.e. deleting an element from one list and adding it to another list does not violate the space limitation.

public <X> void interleaveLists(List<X> first, List<X> second)
{
    ListIterator<X> firstIt = first.listIterator();
    ListIterator<X> secondIt = second.listIterator();
    while(secondIt.hasNext()) {
        fistIt.next();
        firstIt.add(secondIt.next());
        secondIt.remove();
    }
}

This method works for any pair of lists, but is only O(n) for linked lists.

For a custom linked list where we can modify the pointers, we don't have to rely on the garbage collector, we would simply change the nodes. Here for a singly-linked list:

public void interleaveLinkedLists(Node<X> firstList, Node<X> secondList) {
    while(secondList != null) {
        Node<X> nextFirst = firstList.next;
        Node<X> nextSecond = secondList.next;
        firstList.next = secondList;
        secondList.next = nextFirst;
        firstList = nextFirst;
        secondList = nextSecond;
    }
}

For a doubly-linked list, we would also have to adapt the prev-pointers.

Here the wrapping variant mentioned in the first paragraph:

public List<X> interleaveLists(final List<X> first, final List<X> second)
{
   if (first.size() != second.size())
      throw new IllegalArgumentException();
   return new AbstractList<X>() {
      public int size() {
         return 2 * first.size();
      }
      public X get(int index) {
         return index % 2 == 0 ? first.get(index / 2) : second.get(index / 2);
      }
      // if necessary, add a similar set() method.  add/remove are not sensible here.
   };
}

This is actually O(1) in time, too.

I've done up a small solution going on the assumption that you are talking about using the ArrayList (see my comment on the question). I may be oversimplifying the problem based on some of the responses here, but here goes anyway.

The below example takes a and b both of type ArrayList<Integer> and interleaves them by inserting b[0] after a[0], b[1] after a[1] etc. This snippet of course naively assumes that a and b are of the same size as per your Edit v1.0. It also does not create a new ArrayList as per your Edit v2.0.

//a and b are of type ArrayList<Integer>
for (int i = a.size(); i > 0; i--)
{
    a.add(i, b.get(i - 1));
}

No matter what happens if you are combining the ArrayLists you're going to have twice the size.