AI代码解释 publicclassSolution{publicListNodereverseBetween(ListNode head,int m,int n){if(m==n||head==null||head.next==null){returnhead;}ListNode pre=newListNode(0);pre.next=head;ListNode Mpre=pre;ListNode NodeM=head;ListNode NodeN=head;int mNum=1;int nNum=1;while(mNum<m&&NodeM!=nu...
ListNode* reverseBetween(ListNode* head, int m, int n) { if(m==n) return head; auto dummy=new ListNode(-1); dummy->next=head; auto a=dummy,d=dummy; for(int i=0;i<m-1;i++) a=a->next; for(int i=0;i<n;i++) d=d->next; auto b=a->next,c=d->next; for(auto p=...
(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]...
(2)对于reverseList(3)这个情况,此时head为3,head->next为4,此时链表情况如下:1->2->3->4<-5head->next->next=head这一操作后,链表变为:然后,head->next=NULL,即1->2->3<-4<-5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 /** * Definition for singly-linked list. *...
Can you solve this real interview question? Reverse Linked List - Given the head of a singly linked list, reverse the list, and return the reversed list. Example 1: [https://assets.leetcode.com/uploads/2021/02/19/rev1ex1.jpg] Input: head = [1,2,3,
# 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]: # 定义一个哨兵结点,方便后续处理 ...
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(保持下个节点信息不丢) head.next = dummy.next # dummy.next是Null,所以这样head.next就成为了Null ...
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: Givenm,nsatisfy the following condition: 1≤m≤n≤ length of list. ...
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, return1->4->3->2->5->NULL. Note: Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length of list. ...
Reverse a singly linked list. 翻转一个单链表,如:1->2 输出 2->1;1->2->3 输出3->2->1。 题目解析: 本人真的比较笨啊!首先想到的方法就是通过判断链尾是否存在,再新建一个链表,每次移动head的链尾元素,并删除head链表中的元素,一个字“蠢”,但好歹AC且巩固了链表基础知识。你可能遇见的错误包括:...