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:...
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]: ...
# 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...
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 without using extra space? Solution......
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?
publicListNodereverseList(ListNode head){ListNodeprev=null;ListNodecurr=head;while(curr !=null) {ListNodenextTemp=curr.next; curr.next = prev; prev = curr; curr = nextTemp; }returnprev; } 二刷,python。 # Definition for singly-linked list.# class ListNode(object):# def __init__(self, ...
C++ program to reverse a single linked list#include<bits/stdc++.h> using namespace std; class node{ public: int data; // data field node *next; }; node* reverse(node* head){ node *next=NULL,*cur=head,*prev=NULL; //initialize the pointers while(cur!=NULL){//loop till the end ...
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 ...
python # 反转单链表 迭代或递归 class ListNode: def __init__(self, val): self.val = val self.head = None class Solution: def reverseList2(self, head: ListNode) -> ListNode: """ 双指针法 :param head: :return: """ # 空表时直接返空 ...