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2-2-4-linked-list, remove tests accept the broken code #167

Open
@obaibula

Description

The next broken code is accepted:

@Override
 public T remove(int index) {
 checkIndex(index, size);
 if (index == 0) {
 return removeHead();
 } else if (index == size - 1) {
 return removeTail();
 } else {
 return removeInBetween(index);
 }
 }
 private T removeInBetween(int index) {
 Node<T> previous = getNodeByIndex(index - 1);
 T removed = previous.next.element;
 previous.next = previous.next.next;
 size--;
 return removed;
 }
 private T removeTail() {
 T removed = tail.element;
 Node<T> preTail = getNodeByIndex(size - 2);
 preTail.next = null;
 tail = preTail;
 size--;
 return removed;
 }
 private T removeHead() {
 T removed = head.element;
 head = head.next;
 size--;
 return removed;
 }

Please consider the following example:

LinkedList<Integer> list = LinkedList.of(10);
 System.out.println(list.remove(0));
 System.out.println(list.size());
 System.out.println(list.getLast());

In this example, we can still get the "last" element even if size == 0.

It's necessary to reassign tail, as shown below:

private T removeHead() {
 T removed = head.element;
 head = head.next;
 size--;
 if(head == null){
 tail = null;
 }
 return removed;
 }

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