Python获取单向链表倒数第K个节点值的实现方法

#初始化链表的结点
class Node():
  def __init__(self,item):
    self.item = item
    self.next = None
#传入头结点，获取整个链表的长度
def length(headNode):
  if headNode == None:
    return None
  count = 0
  currentNode =headNode
  #尝试了一下带有环的链表，计算长度是否会死循环，确实如此，故加上了count限制 = =||
  while currentNode != None and count <=1000:
    count+=1
    currentNode = currentNode.next
  return count
#获取倒数第K个结点的值，传入头结点和k值
def findrKnode(head,k):
  if head == None:
    return None
  #如果长度小于倒数第K个值,则返回通知没有这么长
  elif length(head)<k:
    print("链表长度没有倒数第"+str(k)+"数")
    return None
  else:
    #设置两个针，一个快，一个慢，都指向头结点
    fastPr = head
    lowPr = head
    count = 0
    #让fastPr先走k个长度
    while fastPr!=None and count<k:
      count+=1
      fastPr = fastPr.next
    #此时fastPr和lowPr同速前进，当fastPr走到尾部，lowPr此处的值正好为倒数的k值
    while fastPr !=None:
      fastPr = fastPr.next
      lowPr = lowPr.next
    return lowPr
if __name__ == "__main__":
  node1 = Node(1)
  node2 = Node(2)
  node3 = Node(3)
  node4 = Node(4)
  node5 = Node(5)
  node6 = Node(6)
  node7 = Node(7)
  node8 = Node(8)
  node9 = Node(9)
  node10 = Node(10)
  node1.next = node2
  node2.next = node3
  node3.next = node4
  node4.next = node5
  node5.next = node6
  node6.next = node7
  node7.next = node8
  node8.next = node9
  node9.next = node10
  print(findrKnode(node1,5).item)