leetocde 876: Find the middle element of linked list

program10.py

# 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:
        fst=head
        slw=head
        while(fst is not None and fst.next is not None):
            fst=fst.next.next
            slw=slw.next
        return slw