Mark Miyashita

public void append(SinglyLinkedList otherList) {

}

Here’s the solution I came up with:

/*
 * This is the append method of the SinglyLinkedList class.
 */
public void append(SinglyLinkedList otherList) {
  if (head == null) {
    head = otherList.head;
  } else {
    SinglyLinkedListNode current = head;
    while (current.next != null) {
      current = current.next;
    }
    current.next = otherList.head;
  }
}

This implementation uses a while loop, however you can also solve this recursively. Basically, we are first checking if we have a head. If we don’t then we simply set our head to point to the other list’s head. If we do have a head, then we iterate over the list until we reach the last item. Then we set the last item’s next to the otherList’s head. The result is a mutated list with the contents of otherList appended to the end.

To check our work, we wrote the following test cases (Note: these are by no means complete):

public static void main(String[] args) {
  Object[] firstListElements = {1, 2, 3, 4, 5};
  Object[] secondListElements = {6, 7, 8, 9, 10};
  SinglyLinkedList p = new SinglyLinkedList(firstListElements);
  SinglyLinkedList q = new SinglyLinkedList(secondListElements);
  System.out.println(p.toString());
  p.append(q);
  System.out.println(p.toString());
  Object[] newElements = {1, 2, 3, 4};
  Object[] noElements = {};
  SinglyLinkedList e = new SinglyLinkedList(newElements);
  SinglyLinkedList f = new SinglyLinkedList(noElements);
  System.out.println(e.toString());
  e.append(f);
  System.out.println(e.toString());
  System.out.println(f.toString());
  f.append(e);
  System.out.println(f.toString());
}

Which should output:

1 2 3 4 5
1 2 3 4 5 6 7 8 9 10
1 2 3 4
1 2 3 4

1 2 3 4