Let's write a `removeAt` method that removes the `element` at a given `index`. The method should be called `removeAt(index)`. To remove an `element` at a certain `index`, we'll need to keep a running count of each node as we move along the linked list.
A common technique used to iterate through the elements of a linked list involves a <dfn>'runner'</dfn>, or sentinel, that 'points' at the nodes that your code is comparing. In our case, starting at the `head` of our list, we start with a `currentIndex` variable that starts at `0`. The `currentIndex` should increment by one for each node we pass.
Just like our `remove(element)` method, which [we covered in a previous lesson](/learn/coding-interview-prep/data-structures/remove-elements-from-a-linked-list), we need to be careful not to orphan the rest of our list when we remove the node in our `removeAt(index)` method. We keep our nodes contiguous by making sure that the node that has reference to the removed node has a reference to the next node.
# --instructions--
Write a `removeAt(index)` method that removes and returns a node at a given `index`. The method should return `null` if the given `index` is either negative, or greater than or equal to the `length` of the linked list.
**Note:** Remember to keep count of the `currentIndex`.
# --hints--
Your `LinkedList` class should have a `removeAt` method.
When only one element is present in the linked list, your `removeAt` method should remove and return the element at specified index, and reduce the length of the linked list.
Your `removeAt` method should return `null` and the linked list should not change if the given index is greater than or equal to the `length` of the list.