开始处理每一个节点ListNode second = first.next;//先处理第一个节点first,所以需要一个指针来存储first的后继first.next = reverseHead;//将first放到新链表头节点的头部reverseHead = first;//移动新链表的头指针,让它始终指向新链表头部first = second;//继续处理原链表的节点,即之前指针存放的后继...
代码: /*** Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * }*/classSolution {publicListNode reverseBetween(ListNode head,intm,intn) {if(m==n)returnhead; ListNode startNode= head;//记录m-1的点ListNode pr...
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...
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. 迭代法 复杂度 ...
Reverse a singly linked list. click to show more hints. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 既然问了能否iteratively or recursively, 那就both把. iterative 解法: 总结就是得到下一个节点,更改当前节点指向,将指针往下移动,直到过完整个linked...
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""" ...
public ListNode reverseList(ListNode head) { if (head == null || head.next == null){ return head; } ListNode p = reverseList(head.next); //重点 head.next.next = head; head.next = null; return p; } 总结 三种解法中其实利用栈来反转是最简单的,其次是迭代,最难的是递归,递归可以想成...
Reverse a singly linked list. 反转链表,我这里是采用头插法来实现反转链表。 代码如下: /*class ListNode { int val; ListNode next; ListNode(int x) { val = x; } } */ public class Solution { public ListNode reverseList(ListNode head) ...
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 ...
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,