AcWing 33. 链表中倒数第k个节点-python-递归
原题链接
简单
作者:
acwing_827
,
2019-09-04 11:07:16
,
所有人可见
,
阅读 1282
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
#递归返回
class Solution(object):
p = 0
def findKthToTail(self, pListHead, k):
"""
:type pListHead: ListNode
:type k: int
:rtype: ListNode
"""
if not pListHead:
return None
node = self.findKthToTail(pListHead.next, k)
if node:
return node
self.p+=1
if self.p == k:
return pListHead
return None