deleteAtPosition<T> function

LinkedListNode<T>? deleteAtPosition<T>(LinkedListNode<T>? head, int position)

Deletes the node at the specified position

head - The head of the linked list position - The position of the node to delete (0-indexed) Returns the new head of the linked list

Time Complexity: O(n) where n is the position Space Complexity: O(1)

Implementation

LinkedListNode<T>? deleteAtPosition<T>(LinkedListNode<T>? head, int position) {
  // Handle empty list
  if (head == null) return null;

  // Handle deletion of the first node
  if (position == 0) {
    return head.next;
  }

  // Handle invalid position
  if (position < 0) return head;

  LinkedListNode<T>? current = head;
  int currentPosition = 0;

  // Traverse to the position before the node to delete
  while (current != null && currentPosition < position - 1) {
    current = current.next;
    currentPosition++;
  }

  // If we reached the end or the next node doesn't exist, return original head
  if (current == null || current.next == null) return head;

  // Delete the node by updating the next pointer
  current.next = current.next!.next;

  return head;
}