【剑指offer】面试题 24：反转链表

题目描述

输入一个链表，反转链表后，输出链表的所有元素。

时间限制：1秒 空间限制：32768K 热度指数：137078

本题知识点： 链表

思路

pHead始终指向要反转的节点，last 指向反转后的首节点，

每反转一个节点，把 pHead 节点的下一个节点指向 last，

last 指向 pHead 成为反转后的首节点，再把 pHead 向前移动一个节点直至None结束；

参考代码

版本一

# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
# 返回ListNode
def ReverseList(self, pHead):
# write code here
if not pHead or not pHead.next:
return pHead

last = None

while pHead:
tmp = pHead.next
pHead.next = last
last = pHead
pHead = tmp

return last

版本二

# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
# 返回ListNode
def ReverseList(self, pHead):
# write code here
pReversedHead = None
pNode = pHead
pPrev = None
while pNode != None:
pNext = pNode.next

if pNext == None:
pReversedHead = pNode

pNode.next = pPrev
pPrev = pNode
pNode = pNext
return pReversedHead

def ReverseListRec(self, pHead):
if not pHead or not pHead.next:
return pHead
else:
pReversedHead = self.ReversedList(pHead.next)
pHead.next.next = pHead
pHead.next = None
return pReversedHead