【剑指Offer】15.反转链表（Python实现）

题目描述

输入一个链表，反转链表后，输出新链表的表头。

解法一：循环法

# -*- 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 pHead is None:
            return pHead
        last = None  #指向上一个节点
        while pHead:
            # 先用tmp保存pHead的下一个节点的信息，
            # 保证单链表不会因为失去pHead节点的next而就此断裂
            tmp = pHead.next
            # 保存完next，就可以让pHead的next指向last了
            pHead.next = last
            # 让last，pHead依次向后移动一个节点，继续下一次的指针反转
            last = pHead
            pHead = tmp
        return lastleetcode

解法二：递归法

# 递归
# -*- 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
        else:
            newHead = self.ReverseList(pHead.next)
            pHead.next.next=pHead
            pHead.next=None
            return newHead