链表倒数第k个节点 

 1 # -*- coding:utf-8 -*-
 2 # class ListNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.next = None
 6 
 7 """双指针的关键是如何保持指针的距离为k-1"""
 8 class Solution:
 9     def FindKthToTail(self, head, k):
10         # write code here
11         if head == None or k==0:
12             return None
13         phead = head     
14         pbehind = head
15         for i in range(k-1): #range的数字之差代表元素个数
16             if phead.next == None:
17                 return None
18             else:
19                 phead = phead.next
20         while phead.next != None: #倒数第二个节点依然满足，因此最终的phead是最后一个，最终的pbehind就是倒数第k个
21             phead = phead.next
22             pbehind = pbehind.next
23         return pbehind
24 
25 
26 """直接用列表实现"""
27 # -*- coding:utf-8 -*-
28 # class ListNode:
29 #     def __init__(self, x):
30 #         self.val = x
31 #         self.next = None
32 
33 class Solution:
34     def FindKthToTail(self, head, k):
35         # write code here
36         l = []
37         while head:
38             l.append(head)
39             head = head.next
40         if len(l) < k or k < 1:
41             return None
42         return l[-k]

 

posted @ 2019-05-06 18:34  Parallax  阅读(89)  评论(0)    收藏  举报