def linked_list_middle(head):
  p1, p2 = head, head
  while p2 != None:
    p2 = p2.next
    if p2 == None:
      break
    p1 = p1.next
    p2 = p2.next
  return p1

#Challenge

Given a non-empty, singly, linked list, return a pointer to the middle of the list. If the list contains an even number of elements, return the one closer to the end.

#Solution

Maintain two pointers with two different step widths. For every step forward the first pointer takes, make the other take 2. Once the second pointer has reached the end of the array, the first should be at the middle of it.

#Linked List Implementation

class Node:
  def __init__(self, value, next=None):
    self.value = value
    self.next = next

  @classmethod
  def from_list(cls, lst):
    parent = None
    for elem in reversed(lst):
      parent = cls(elem, next=parent)
    return parent