removeDuplicateNodeFromLinkedList

/**
 * Interview Question:
 * Given the head of a sorted singly linked list, remove duplicate values in
 * place and return the original head.
 *
 * Input:
 * 1 -> 1 -> 3 -> 4 -> 4 -> 4 -> 5 -> 6 -> 6
 *
 * Output:
 * 1 -> 3 -> 4 -> 5 -> 6
 *
 * Requirements:
 * - The linked list is already sorted
 * - Remove duplicate nodes, not just duplicate values
 * - Keep one copy of each value
 * - Modify the list in place
 * - Return the head of the cleaned list
 *
 * Answer:
 * Because the list is sorted, duplicate values are adjacent. Walk through the
 * list with one pointer. If current.value equals current.next.value, skip the
 * next node by setting current.next = current.next.next. Otherwise, move current
 * forward.
 *
 * Complexity:
 * O(n) time because each node is visited at most once.
 * O(1) space because the list is updated in place.
 *
 * Node shape used here:
 * class LinkedList {
 *   constructor(value) {
 *     this.value = value;
 *     this.next = null;
 *   }
 * }
 *
 * If your node uses `val` instead of `value`, replace current.value with
 * current.val.
 *
 * @param {{ value: number, next: object | null } | null} head
 * @return {{ value: number, next: object | null } | null}
 */
export default function removeDuplicateNodesFromLinkedList(head) {
  let current = head;

  while (current !== null && current.next !== null) {
    if (current.value === current.next.value) {
      current.next = current.next.next;
    } else {
      current = current.next;
    }
  }

  return head;
}