}ListNodep=reverseList(head.next); head.next.next = head; head.next =null;returnp; } } Python 实现 # Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = NoneclassSolution:defreverseList(self, head):""" :type head: ListNode :...
(参考视频讲解:Leetcode力扣|206反转链表|递归|reverse linked list_哔哩哔哩_bilibili) # 定义一个链表节点类classListNode:def__init__(self,val=0,next=None):# 初始化函数self.val=val# 节点的值self.next=next# 指向下一个节点的指针# 将给出的数组转换为链表deflinkedlist(list):head=ListNode(list[0]...
代码示例1 (python 轮回版) # Definition for singly-linked list.# class ListNode(object):# def __init__(self, x):# self.val = x# self.next = NoneclassSolution(object):defreverseList(self, head):""" :type head: ListNode :rtype: ListNode """cur_node = head pre_node =Nonewhilecur...
Leetcode 92题反转链表 II(Reverse Linked List II) LeetCode 206题 反转链表(Reverse Linked List) 题目链接 https://leetcode-cn.com/problems/reverse-linked-list-ii/ 题目描述 反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。 说明: 1 ≤ m ≤ n ≤ 链表长度。 示例: 输入: 1->2->3->4-...
Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL. Note: Given m, n satisfy the following condition: ...
LeetCode上第206号问题:Reverse Linked List 反转一个单链表。 示例: 输入: 1->2->3->4->5->NULL 输出: 5->4->3->2->1->NULL 进阶: 你可以迭代或递归地反转链表。你能否用两种方法解决这道题? 思路 设置三个节点pre、cur、next (1)每次查看cur节点是否为NULL,如果是,则结束循环,获得结果 ...
def reverseList(self, head): """ :type head: ListNode :rtype: ListNode """ dummy = ListNode(None) while head: # 终止条件是head=Null nextnode = head.next # nextnode是head后面的节点 head.next = dummy # dummy.next是Null,所以这样head.next就成为了Null ...
LeetCode 206. 反转链表(Reverse Linked List) 示例: 输入:1->2->3->4->5->NULL输出:5->4->3->2->1->NULL 切题 一、Clarification 只需注意为空链表的情况 二、Possible Solution 1、迭代 2、递归 可利用哨兵简化实现难度 Python3 实现
LeetCode Reverse a singly linked list. Example: Input:1->2->3->4->5->NULL Output:5->4->3->2->1->NULL Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both? 问题 力扣 反转一个单链表。
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