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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. 




class Solution {
    public ListNode deleteMiddle(ListNode head) {
        
        if (head == null ||head.next == null) {
            return null;
        }

        int count = 0;
        ListNode node = head;

        while (node != null) {
            count++;
            node = node.next;
        }

        int mid = count / 2;
        int temp = 0;

        node = head;

        while (node != null) {
            if (temp == mid - 1) {
                node.next = node.next.next;
                break;
            }
            node = node.next;
            temp++;
        }

        return head;
    }
}