题解 | #输出单向链表中倒数第k个结点#

import java.util.Scanner;
class ListNode {
    int m_nKey;
    ListNode m_pNext;
    ListNode(int val) {
        m_nKey = val;
        m_pNext = null;
    }
}
public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNext()) {
            int n = scanner.nextInt(); // 链表结点个数
            ListNode head = new ListNode(scanner.nextInt()); // 构建头结点
            ListNode cur = head;
            for (int i = 1; i < n; i++) {
                cur.m_pNext = new ListNode(scanner.nextInt());
                cur = cur.m_pNext;
            }
            int k = scanner.nextInt(); // 倒数第k个结点

            ListNode result = findKthToTail(head, k);
            System.out.println(result != null ? result.m_nKey : "空指针异常");
        }
        scanner.close();
    }
    // 寻找倒数第k个结点
    public static ListNode findKthToTail(ListNode head, int k) {
        if (head == null || k <= 0) 
            return null;
        ListNode slow = head;
        ListNode fast = head;
        // 让快指针先走k步
        for (int i = 0; i < k; i++) {
            if (fast == null) 
                return null; // 如果链表长度小于k，返回null
            fast = fast.m_pNext;
        }
        // 快慢指针一起走，直到快指针到达链表尾部
        while (fast != null) {
            slow = slow.m_pNext;
            fast = fast.m_pNext;
        }
        return slow; // 返回倒数第k个结点
    }
}