---
tags: data_structure_python
---# Middle of the Linked List <img src="https://img.shields.io/badge/-easy-brightgreen">

Given a non-empty, singly linked list with head node head, return a middle node of linked list.

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

<ins>**Example 1**:</ins>

Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5])
The returned node has value 3.  (The judge's serialization of this node is [3,4,5]).
Note that we returned a ListNode object ans, such that:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.

<ins>**Example 2**:</ins>

Input: [1,2,3,4,5,6]
Output: Node 4 from this list (Serialization: [4,5,6])
Since the list has two middle nodes with values 3 and 4, we return the second one.

**Note:**

- The number of nodes in the given list will be between 1 and 100.

# Solution

### Solution 1: First approach

# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def middleNode(self, head: ListNode) -> ListNode:
        # O(n) time complexity.
        # O(1) space complexity.
        count = 0
        pointer2 = head
        while pointer2 != None:
            count += 1
            pointer2 = pointer2.next
        
        mid = count//2
        while mid > 0:
            head = head.next
            mid -= 1
            
        return head

### Solution 2: Second approach (Two pointers)

# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def middleNode(self, head: ListNode) -> ListNode:
        fast = head
        slow = head
        while fast and fast.next:
            fast, slow = fast.next.next, slow.next
        return slow