单向链表反向输出,不用中间变量操作.

讲一下基本思想:
单向链表要想反向输出,而不借助临时变量,那就是栈这个数据结构->递归一下就出现了.

import lombok.AllArgsConstructor;
import lombok.Data;
import lombok.NoArgsConstructor;
import lombok.experimental.Accessors;

import java.util.Objects;

@Data
@AllArgsConstructor
@NoArgsConstructor
@Accessors(chain = true)
class Node {
    private Integer data;
    private Node next;
}
/**
 * @author 陈英俊
 * @date 2020/11/30 17:46
 */
public class SingleLinkedListDemo {
    public static void main(String[] args) {
        Node head = new Node(0, null);
        for (int i = 0; i < 10; i++) {
            initLinkedList(head, new Node(i, null));
        }
        Node tmp = head;
        while (tmp != null) {
            System.out.print(tmp.getData() + "->");
            tmp = tmp.getNext();
        }
        System.out.println();
        //单链表反向输出
        printLinkedList(head);
    }
    /**
     * 单向链表反向输出
     *
     * @param head 头结点
     */
    private static void printLinkedList(Node head) {
        if (!Objects.isNull(head.getNext())) {
            printLinkedList(head.getNext());
        }
        System.out.print(head.getData() + "->");
        return;
    }

     /**
     * 链表新增结点(尾插法)
     *
     * @param head 头结点
     * @param add 新增结点
     */
    private static void initLinkedList(Node head,Node add) {
        Node tmp = head.getNext();
        Node var = head;
        while (tmp != null) {
            var = tmp;
            tmp = tmp.getNext();
        }
        var.setNext(add);
    }
}

• 输出结果:

0->1->2->3->4->5->6->7->8->9->10->
10->9->8->7->6->5->4->3->2->1->0->
Process finished with exit code 0