题解 | #输出单向链表中倒数第k个结点#

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while(in.hasNextInt()){
           int n = in.nextInt();
           int i = 1;
           ListNode head = new ListNode(-1);//创建一个头结点
           ListNode headNode = head;
           //创建链表
           for (int j = 0; j < n; j++) {
               ListNode node = new ListNode(in.nextInt());
               head.next = node;
               head = node;
           }
           int num = in.nextInt();
           //定义一前一后两指针，让前面的指针与后面的指针相差指定的num
           //现在head就在我们最后一个节点,但是haedNode是第一个我们的头结点
           ListNode first = headNode;
           for (int j = 0; j < n-num+1; j++) {
               first = first.next;
           }
           System.out.println(first.val);
       }
        in.close();
    }
}
class ListNode {
    int val;
    ListNode next;

    public ListNode(int val) {
        this.val = val;
    }

    public ListNode(int val, ListNode next) {
        this.val = val;
        this.next = next;
    }
}