# Delete the Middle Node of a Linked List

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`

### Solution:

```# 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]:
dummy = ListNode()
slow = fast = dummy
while fast.next and fast.next.next:
slow = slow.next
fast = fast.next.next
slow.next = slow.next.next
return dummy.next
```

Ashwini Verma on Jan 30, 2022 at 11:01 am

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