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
一时之间没有想到怎样做reverse比较好,参考了一下网上的思路,发现这样做比较好:还是要用Runner Technique,还是要用Dummy Node;两个指针: npointer指到n的位置,mpointer指到m的前一位;每一次把mpointer后一位的元素放到npointer的后一位:mpointer.next.next = npointer.next;直到mpointer.next = npointer为止(m与...
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...
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...
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・...
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: ...
/** * Definition for singly-linked list. * type ListNode struct { * Val int * Next *ListNode * } */ func reverseBetween(head *ListNode, left int, right int) *ListNode { // 定义一个哨兵结点,方便后续处理 headPre := &ListNode{ Val: 0, Next: head } // 先找到第 m 个结点的前一...
[LeetCode]Reverse Linked List II Question 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....
来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/reverse-linked-list-ii/ 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。解法一：利用栈首先，如果head为null或者head只有一个节点，直接返回head。否则， 声明一个新的头节点newHead，声明一个栈reverseNodes用来放left和...
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。