题解 | #输出单向链表中倒数第k个结点#

import java.util.*;
public class Main {
    public static class ListNode {
        public int val;
        public ListNode next;
        public ListNode(int val) {
            this.val = val;
        }
    }
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        while (scan.hasNext()) {
            int n = Integer.valueOf(scan.nextLine().trim());
            String[] nums = scan.nextLine().split(" ");
            int k = Integer.valueOf(scan.nextLine().trim());
            ListNode head = createList(nums);
            find(head, k);
        }
    }
    // 自定义一个函数，实现链表的创建
    public static ListNode createList(String[] nums) {
        ListNode head = new ListNode(1);
        ListNode node = new ListNode(1);
        for (int i = 0; i < nums.length; i++) {
            if (i == 0) {
                head = new ListNode(Integer.valueOf(nums[i]));
                node = head;
                continue;
            }
            node.next = new ListNode(Integer.valueOf(nums[i]));
            node = node.next;
        }
        node.next = null;
        return head;
    }
    // 自定义一个函数，找到链表的倒数第 k 个节点
    public static void find(ListNode head, int k) {
        ListNode node = head;
        int len = 0;
        while (null != node) {
            len++;
            node = node.next;
        }
        int ans = 0;
        node = head;
        for (int i = 1; i <= len - k + 1; i++) {
            ans = node.val;
            node = node.next;
        }
        System.out.println(ans);
    }
}