How does add(index, element) method work behind the scenes using LinkedList?

Question

When adding an element at a certain index using the add(index, element) method in an ArrayList, it places the element at that index, while all the other elements change their indexes by 1 (they move in memory). That's why an ArrayList has the complexity O(n) when adding an element at a certain position.

In case of a doubly LinkedList, I know that the elements have pointers to the previous element and also for the next one.

My question is, when using the add(index, element) method related to a LinkedList, what will actually happen behind the scenes? I know that using the LinkedList the rest of the elements don't move in memory, so how come they can still be placed at a certain index without moving in memory?

The indices on a LinkedList don't correspond to contiguous memory locations. Instead, the get() method requires a list traversal to access the correct index. This is why ArrayList::get has a complexity of O(1) (it can just jump to the correct memory address), but LinkedList::get has a complexity of O(n). — Jordan
– Jordan, Commented Sep 23, 2019 at 15:23

Tim Biegeleisen · Accepted Answer · 2019-09-23 15:38:08Z

2

Adding a new element to a linked list at a specified index requires walking down the list (from either the head or tail), and then splicing in the new element. The source code for LinkedList#add(int index, E element) reveals as much:

public void add(int index, E element) {
    checkPositionIndex(index);

    if (index == size)
        linkLast(element);
    else
        linkBefore(element, node(index));
}

Should the index be pointing to the final item in the list, it simply appends a new node to the end. Otherwise, it calls linkBefore(), which does a bit of splicing work (I won't bother to include its source code as well).

Note here that adding a new node to a linked list does not necessarily involve moving anything in the already existing list. Rather, it mostly involves moving around references in the background.

edited Sep 23, 2019 at 15:38

answered Sep 23, 2019 at 15:23

Tim Biegeleisen

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2 Comments

Tim Biegeleisen Over a year ago

@Michael I actually was worried someone would call me out on that. I changed my language a bit.

Michael Over a year ago

Welcome to pedant country

Jacob G. · Accepted Answer · 2019-09-23 15:25:03Z

The implementation of add(index, element) looks like this:

public void add(int index, E element) {
    checkPositionIndex(index);

    if (index == size)
        linkLast(element);
    else
        linkBefore(element, node(index));
}

If you're adding an element to the tail of the LinkedList, the linkLast method can be executed in constant time; the LinkedList always has direct access to its last element, and no traversing is required.

Otherwise, the node method is what's expensive, as a traversal through at most half of the list is required:

Node<E> node(int index) {
    // assert isElementIndex(index);

    if (index < (size >> 1)) {
        Node<E> x = first;
        for (int i = 0; i < index; i++)
            x = x.next;
        return x;
    } else {
        Node<E> x = last;
        for (int i = size - 1; i > index; i--)
            x = x.prev;
        return x;
    }
}

The elements aren't required to move in memory because they each refer to their previous and next nodes in the LinkedList, as you can see below in linkBefore:

void linkBefore(E e, Node<E> succ) {
    // assert succ != null;
    final Node<E> pred = succ.prev;
    final Node<E> newNode = new Node<>(pred, e, succ);
    succ.prev = newNode;
    if (pred == null)
        first = newNode;
    else
        pred.next = newNode;
    size++;
    modCount++;
}

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