Java写题解的第5天 | #输出单向链表中倒数第k个结点#

解题思路：使用快慢指针，令快指针先指向头节点，再走k个位置，再令慢指针指向头节点，同时先后移动直到快指针指向null；
注：由于本题的输入已知了链表的长度，因此也可以在建立好链表之后直接移动到n-k的位置；

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNext()) {
            int len = sc.nextInt();
            ListNode pHead = new ListNode(sc.nextInt());
            ListNode build = pHead;
            for (int i = 1; i < len; i++) {
                ListNode newNode = new ListNode(sc.nextInt());
                build.next = newNode;
                build = newNode;
            }
            int k = sc.nextInt();
            if (k == 0 || k > len) {
                System.out.println(0);
            } else {
                ListNode fast = pHead;
                for (int i = 0; i < k; i++) {
                    fast = fast.next;
                }
                ListNode slow = pHead;
                while (fast != null) {
                    slow = slow.next;
                    fast = fast.next;
                }
                System.out.println(slow.val);
            }

        }
    }
}

class ListNode {
    int val;
    ListNode next;

    public ListNode() {

    }

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

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