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# Delete the Middle Node of a Linked List

The middle node of a linked list of size n is the βn / 2βth node from the start using 0-based indexing, where βxβ 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:
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:
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:
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.

``````/**
* function ListNode(val, next) {
*     this.val = (val===undefined ? 0 : val)
*     this.next = (next===undefined ? null : next)
* }
*/
/**
* @return {ListNode}
*/
let dummy = new ListNode(); //to deal with edge case that we need to delete the head node

let prev = dummy;

//both pointers travel
while(fast && fast.next){
fast = fast.next.next;
prev = slow;
slow = slow.next;
}

//delete the node
while(slow.next){
slow.val = slow.next.val;
prev = slow;
slow = slow.next;
}

prev.next = null;
return dummy.next;
};
``````