# [leetcode] 2095. Delete the Middle Node of a Linked List _ Algorithm Problem Solve for python

## 1. Problem

2095. Delete the Middle Node of a 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: 
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, 10^5].
• 1 <= Node.val <= 10^5

## 2. Solution

I solved this problem like this.

• Check node count.
• Check delete node idx and delete.
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
def deleteMiddle(self, head: Optional[ListNode]) -> Optional[ListNode]:
cnt = 0
while node:
node = node.next
cnt += 1

if cnt == 1:
return

del_idx = cnt // 2
idx = 0