某公司网站源码体验营销案例

概述

本文介绍了一种使用 ArrayList 和 Collections.reverse() 方法来实现链表反转的解决方案，虽然这种方法在空间复杂度上不是最优的，但代码简洁易懂。

实现思路

1. 遍历原链表，将所有节点的值收集到 ArrayList 中
2. 使用 Collections.reverse() 方法反转 ArrayList
3. 根据反转后的值列表重新构建链表

源码实现

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;public class ReverseNode {public ListNode reverseList(ListNode head) {if (head == null || head.next == null) {return head;}// 定义一个收集值的数组List<Integer> valList = new ArrayList<>();while (head.next != null) {valList.add(head.val);// 移动链表head = head.next;}// 加上最后一个节点的值valList.add(head.val);// 反转收集值Collections.reverse(valList);ListNode cursorNode = new ListNode();for (int index = 0; index < valList.size(); index++) {Integer val = valList.get(index);if (index == 0) {// 初始化头节点head = new ListNode(val);cursorNode = head;continue;}cursorNode.next = new ListNode(val);// 移动游标节点cursorNode = cursorNode.next;}return head;}static class ListNode {int val;ListNode next;ListNode() {}ListNode(int val) {this.val = val;}ListNode(int val, ListNode next) {this.val = val;this.next = next;}}public static void main(String[] args) {ReverseNode reverseNode = new ReverseNode();// 测试用例1：正常链表 1->2->3->4->5ListNode head1 = new ListNode(1);head1.next = new ListNode(2);head1.next.next = new ListNode(3);head1.next.next.next = new ListNode(4);head1.next.next.next.next = new ListNode(5);System.out.println("原链表1：");printList(head1);ListNode reversed1 = reverseNode.reverseList(head1);System.out.println("反转后：");printList(reversed1);System.out.println();// 测试用例2：单节点链表ListNode head2 = new ListNode(42);System.out.println("原链表2：");printList(head2);ListNode reversed2 = reverseNode.reverseList(head2);System.out.println("反转后：");printList(reversed2);System.out.println();// 测试用例3：空链表ListNode head3 = null;System.out.println("原链表3：");printList(head3);ListNode reversed3 = reverseNode.reverseList(head3);System.out.println("反转后：");printList(reversed3);System.out.println();// 测试用例4：两个节点的链表ListNode head4 = new ListNode(10);head4.next = new ListNode(20);System.out.println("原链表4：");printList(head4);ListNode reversed4 = reverseNode.reverseList(head4);System.out.println("反转后：");printList(reversed4);}private static void printList(ListNode head) {if (head == null) {System.out.println("null");return;}ListNode current = head;while (current != null) {System.out.print(current.val);if (current.next != null) {System.out.print(" -> ");}current = current.next;}System.out.println();}
}