python实现获取单向链表倒数第k个结点的值示例

本文实例讲述了python实现获取单向链表倒数第k个结点的值。分享给大家供大家参考，具体

本文实例讲述了python实现获取单向链表倒数第k个结点的值。分享给大家供大家参考，具体如下：

#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)

运行结果：

6

python 单向链表 倒数第k个结点的值