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=h
Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Given1->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. Analysis: 这道题还是...
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. 把[m,n]那一段抠出来,reverse之后,再拼回...
publicclassSolution{publicListNodeReverseList(ListNode head){if(head==null)returnnull;//head为当前节点,如果当前节点为空的话,那就什么也不做,直接返回null;ListNode pre=null;ListNode nextnode=null;while(head!=null){nextnode=head.next;head.next=pre;pre=head;head=nextnode;}returnpre;}} Reverse Lin...
next = None #第 4 步:同第 206 题,反转链表的子区间 reverse_linked_list(left_node) #第 5 步:接回到原来的链表中 pre.next = right_node left_node.next = curr return dummy_node.next 3. 解析 处理过206题类似,只不过这个是一个区间,对区间进行处理,5步解决。 发布于 2024-04-22 17:48・...
# 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 ...
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: ...
2. 完成inverse。 【注意】 本题需要用到fake。 【附】 本题与*(重点)[LeetCode]Reverse Nodes in k-Group不同在于本题的length>=n,而那道题未必能。 【代码】 /** * Definition for singly-linked list. * public class ListNode { ...
pre->next = cur->next; cur->next =front; front=cur; } cur= pre->next; pre->next =front; last->next =cur;returndummy->next; } }; 本文转自博客园Grandyang的博客,原文链接:倒置链表之二[LeetCode] Reverse Linked List II,如需转载请自行联系原博主。
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:...