## Question

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:**

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`

## Python Solution

# 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) -> ListNode: f=head s=head while 1: if f.next==None: return s elif f.next.next==None: return s.next f=f.next.next s=s.next