2095. Delete the Middle Node of a Linked List.js

/*

You are given the head of a linked list. Delete the middle node, and return the head of the modified linked list.

The middle node of a linked list of size n is the ⌊n / 2⌋th node from the start using 0-based indexing, where ⌊x⌋ denotes the largest integer less than or equal to x.

For n = 1, 2, 3, 4, and 5, the middle nodes are 0, 1, 1, 2, and 2, respectively.


Example 1:

Input: head = [1,3,4,7,1,2,6]
Output: [1,3,4,1,2,6]
Explanation:
The above figure represents the given linked list. The indices of the nodes are written below.
Since n = 7, node 3 with value 7 is the middle node, which is marked in red.
We return the new list after removing this node. 
Example 2:


Input: head = [1,2,3,4]
Output: [1,2,4]
Explanation:
The above figure represents the given linked list.
For n = 4, node 2 with value 3 is the middle node, which is marked in red.
Example 3:


Input: head = [2,1]
Output: [2]
Explanation:
The above figure represents the given linked list.
For n = 2, node 1 with value 1 is the middle node, which is marked in red.
Node 0 with value 2 is the only node remaining after removing node 1.
 

Constraints:

The number of nodes in the list is in the range [1, 105].
1 <= Node.val <= 105

*/

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var deleteMiddle = function(head) {
  const listSize = (pt, s = 1) => pt.next ? listSize(pt.next, s + 1) : s;

  let curr = head;
  let seenCount = 0;
  let size = listSize(head);
  const midIdx = Math.floor(size / 2);

  // basic checks
  if (size === 0) { return null; }
  if (size === 1) { return null; }
  if (size === 2) {
    head.next = null;
    return head;
  }

  console.log('midIdx', midIdx);

  while (curr) {
    seenCount++;
    if (seenCount === midIdx) {
      curr.next = curr.next.next ?? null;
      curr = curr.next;
    } else {
      curr = curr.next;
    }
  }

  return head;
};