代码:oj测试通过 Runtime: 65 ms 1#Definition for singly-linked list.2#class ListNode:3#def __init__(self, x):4#self.val = x5#self.next = None67classSolution:8#@param head, a ListNode9#@param m, an integer10#@param n, an integer11#@return a ListNode12defreverseBetween(self, head...
王几行xing:【Python-转码刷题】LeetCode 203 移除链表元素 Remove Linked List Elements 提到翻转,熟悉Python的朋友可能马上就想到了列表的 reverse。 1. 读题 2. Python 中的 reverse 方法 list1 = list("abcde") list1.reverse() list1 [Out: ] ['e', 'd', 'c', 'b', 'a'] 2.1 reverse 在...
来自专栏 · python算法题笔记# Definition for singly-linked list.# class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: ''' # 递归 def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: ...
第一种方法:迭代 代码语言:javascript 代码运行次数:0 classListNode(object):def__init__(self,x):self.val=x self.next=NoneclassSolution(object):defreverseList(self,head):""":type head:ListNode:rtype:ListNode""" pre=cur=Noneifhead:pre=head cur=head.next pre.next=Noneelse:returnNonewhilecur:...
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。 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?
Reverse Linked List II 题目大意 翻转指定位置的链表 解题思路 将中间的执行翻转,再将前后接上 代码 迭代 代码语言:javascript 代码运行次数:0 运行 AI代码解释 classSolution(object):# 迭代 defreverseBetween(self,head,m,n):""":type head:ListNode:type m:int:type n:int:rtype:ListNode""" ...
Reverse Linked List II 题目大意 翻转指定位置的链表 解题思路 将中间的执行翻转,再将前后接上 代码 迭代 class Solution(object): # 迭代 def reverseBetween(self, head, m, n): """ :type head: ListNode :type m: int :type n: int :rtype: ListNode ...
链接:https://leetcode-cn.com/problems/reverse-linked-list https://leetcode-cn.com/problems/reverse-linked-list/solution/shi-pin-jiang-jie-die-dai-he-di-gui-hen-hswxy/ python # 反转单链表 迭代或递归 class ListNode: def __init__(self, val): ...
```python class ListNode: def __init__(self, val=0, next=None): self.val = val self.next = next def reverse_linked_list(head): prev = None curr = head while curr: next_node = curr.next curr.next = prev prev = curr curr = next_node return prev # 测试 node1 = ListNode(1)...
python class Solution(object): def reverseList(self, head): """ :type head: ListNode :rtype: ListNode """ p1 = None p2 = head while(p2 is not None): tmp = p2.next p2.next = p1 p1 = p2 p2 = tmp return p1 Reverse Linked List II ...