876_middle-of-the-linked-list.py

# https://leetcode.com/problems/middle-of-the-linked-list/

# 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]:
        if head.next is None:
            return head
        
        slow = head
        fast = head
        
        while fast is not None and fast.next is not None:
            slow = slow.next
            fast = fast.next.next
        
        return slow