2095. Delete the Middle Node of a Linked List

[DNPotapov]

-# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def deleteMiddle(self, head):
        """
        :type head: Optional[ListNode]
        :rtype: Optional[ListNode]
        """
        cur = head
        count = 0 
        while cur:
            count += 1
            cur = cur.next
        count = count // 2
        if count == 0:
            head = None
            return head
        cur = head
        while ((count - 1) > 0):
            cur = cur.next
            count = count - 1
        cur.next = cur.next.next
        return head

[DNPotapov]

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