【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 = Noneclass 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 = Noneclass 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