题解 | #输出单向链表中倒数第k个结点#

import java.util.Scanner;

// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        // 注意 hasNext 和 hasNextLine 的区别
        while (in.hasNext()) { // 注意 while 处理多个 case
            int n = in.nextInt();
            ListNode head = new ListNode(-1);
            ListNode tmp = head;
		  // 构建链表
            for (int i = 0; i < n; i++) {
                ListNode node = new ListNode(in.nextInt());
                tmp.next = node;
                tmp = tmp.next;
            }
            int k = in.nextInt();
		  // 快慢指针
            ListNode fast = head;
            ListNode slow = head;
            // fast 先走k步
            for (int i = 0; i < k; i++) {
                if (fast == null) {
                    fast = null;
                } else {
                    fast = fast.next;
                }
            }
            ListNode ans = head;
		  // 快指针走到末尾，慢指针未知即为倒数第k个元素
            while (fast != null) {
                fast = fast.next;
                slow = slow.next;
            }
            System.out.println(slow.val);
        }
    }


}

class ListNode {
    ListNode next;
    int val;
    ListNode(int val) {
        this.val = val;
        next = null;
    }
}