python实现获取单向链表倒数第k个结点的值示例

#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 lowPrif __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)