题解 | #输出单向链表中倒数第k个结点#

关键在于：了解单项链表结构。

另外有问题需要进一步学习：值传递与引用传递。

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        while (scan.hasNextInt()) {
            int n = scan.nextInt();
            //定义表头
            ListNode headNode = new ListNode(-1);
            ListNode temp = headNode;
            //建立单项链表
            for (int i = 0; i < n; i++) {
                int value = scan.nextInt();
                temp.nextNode = new ListNode(value);
                temp = temp.nextNode;
            }

            //获得倒数k值
            int k = scan.nextInt();
            for (int i = 0; i < n - k + 1;
                    i++) { //倒数第k个，即为顺数第n-k+1个，序号则为n-k
                headNode = headNode.nextNode;
            }
            //输出
            System.out.println(headNode.value);
        }

    }
}

class ListNode {
    Integer value;
    ListNode nextNode;
    ListNode() {};
    ListNode(Integer value) {
        this.value = value;
        this.nextNode = null;
    }
}