Linked Lists:Reverse a doubly linked list ...Reverse Linked List #记录经典问题,方便以后复习。 题目: Reverse a singly linked list. Example: 1、非递归 2、递归 ...leetcode: Linked List Question Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it ...
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:...
reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: if not head: return head p = head q = p.next p.next = None while q: r = q.next q.next = p p = q q = r return p发布于 2024-03-21 10:22・北京 Python Python 入门...
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def reverseBetween(self, head: Optional[ListNode], left: int, right: int) -> Optional[ListNode]: i = 0 p = head p_l = None...
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?
leetcode 【 Reverse Linked List II 】 python 实现 题目: Reverse a linked list from positionmton. Do it in-place and in one-pass. For example: Given1->2->3->4->5->NULL,m= 2 andn= 4, return1->4->3->2->5->NULL. Note:...
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""" ...
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 ...
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```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)...