# 876. Middle of the Linked List

## Easy

***

Given the `head` of a singly linked list, return *the middle node of the linked list*.

If there are two middle nodes, return **the second middle** node.

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/07/23/lc-midlist1.jpg)

```
Input: head = [1,2,3,4,5]
Output: [3,4,5]
Explanation: The middle node of the list is node 3.
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2021/07/23/lc-midlist2.jpg)

```
Input: head = [1,2,3,4,5,6]
Output: [4,5,6]
Explanation: Since the list has two middle nodes with values 3 and 4, we return the second one.
```

**Constraints:**

* The number of nodes in the list is in the range `[1, 100]`.
* `1 <= Node.val <= 100`

```python
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
        ptr1 = head
        ptr2 = head
        while ptr2 and ptr2.next:
            ptr1 = ptr1.next
            ptr2 = ptr2.next.next
        return ptr1
```