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.

```
/**
* 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) {
let fast = head;
let slow = head;
let dummy = new ListNode(); //to deal with edge case that we need to delete the head node
dummy.next = head;
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;
};
```

## Latest comments (0)