---
title: '876. Middle of the Linked List'
description: Given the head of a singly linked list, return the middle node of the linked list
icon: dot
topics:
  - { name: "Linked List", slug: "linked-list" }
  - { name: "Two Pointers", slug: "two-pointers" }
issue: "https://github.com/prdlk/leetcode/issues/82"
companies:
  - { name: "Google", period: "Thirty Days", frequency: 50.2 }
  - { name: "Meta", period: "Three Months", frequency: 49.8 }
  - { name: "Amazon", period: "Three Months", frequency: 39.1 }
  - { name: "IBM", period: "Six Months", frequency: 60.1 }
  - { name: "Microsoft", period: "Six Months", frequency: 37.9 }
  - { name: "Bloomberg", period: "Six Months", frequency: 25.5 }
sidebar:
  label: 'Middle of the Linked List'
  badge: 'Easy'
---

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

### Example 2:
- 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`

## Solution

```py
# 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: ListNode | None) -> ListNode | None:
        slow = fast = head
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next

        return slow
```