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? 递归法 复杂度 时间O(N) 空间 O(N) 递归栈空间 思路 基本递归 代码 public clas
206--Reverse A Singly Linked List package LinedList; public class ReverseASinglyLinkedList { //解法一:迭代。 public ListNode reverseList(ListNode head) { ListNode previous = null; ListNode current = head; while (current != null) { ListNode next = current.next; current.next = previous; previo...
Space O(n), 迭代用了stack一共O(n)大小, n 是原来list的长度。 AC Java: 1/**2* Definition for singly-linked list.3* public class ListNode {4* int val;5* ListNode next;6* ListNode(int x) { val = x; }7* }8*/9publicclassSolution {10publicListNode reverseList(ListNode head) {11/...
206. Reverse Linked List 题目: Reverse a singly linked list. 思路: 递归的方法 代码: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* reverseList(Lis...
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) ...
Reverse Linked List Reverse a singly linked list. Note Create tail = null; Head loop through the list: Store head.next, head points to tail, tail becomes head, head goes to stored head.next; Return tail. Solution public class Solution { ...
https://leetcode-cn.com/problems/reverse-linked-list-ii/ 题目描述 反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。 说明: 1 ≤ m ≤ n ≤ 链表长度。 示例: 输入: 1->2->3->4->5->NULL, m = 2, n = 4 输出: 1->4->3->2->5->NULL ...
网络反转单向链表;翻转单向链表 网络释义
Write a C program to reverse alternate k nodes of a given singly linked list. Sample Solution:C Code:#include<stdio.h> #include <stdlib.h> // Definition for singly-linked list struct Node { int data; struct Node* next; }; // Function to create a new node in the linked list struct...
java: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode reverseKGroup(ListNode head, int k) { if (head == ...