1、Java代码(直接运行版)
package listnodes;
import java.util.Scanner;
//Definition for singly-linked list.
class ListNode {
int val;
ListNode next;
ListNode() {}
ListNode(int val) { this.val = val; }
ListNode(int val, ListNode next) { this.val = val; this.next = next; }
}
public class LinkedListTest {
public static void main(String[] args) {
int[] nums = new int[]{1, 2, 3, 4, 5, 6, 7, 8};
//正序构建链表
ListNode head = new ListNode(-1);
ListNode temp = head;
for (int i = 0; i < nums.length; i++) {
ListNode node = new ListNode(nums[i]);
temp.next = node;
temp = temp.next;
}
head = head.next;//去掉虚拟头节点就是构建好的完整链表
//构建链表后,忘记链表长度,重新遍历链表,找到倒数第k个节点
int k = 4;
ListNode pre = head;
ListNode rear = head;
int count = 1;
while(rear.next != null && count < k){
count++;
rear = rear.next;
}
while(rear.next != null){
rear = rear.next;
pre = pre.next;
}
System.out.println(pre.val);//5
}
}
2、Java代码(牛客提交版)
import java.util.Scanner;
class ListNode {
int val;
ListNode next;
ListNode() {}
ListNode(int val) {
this.val = val;
}
ListNode(int val, ListNode next) {
this.val = val;
this.next = next;
}
}
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while (in.hasNextInt()) {
int n = in.nextInt();
//正序构建链表
ListNode head = new ListNode(-1);
ListNode temp = head;
for (int i = 0; i < n; i++) {
ListNode node = new ListNode(in.nextInt());
temp.next = node;
temp = temp.next;
}
head = head.next;//去掉虚拟头节点就是构建好的完整链表
//构建链表后,忘记链表长度,重新遍历链表,找到倒数第k个节点
int k = in.nextInt();
ListNode pre = head;
ListNode rear = head;
int count = 1;
while (rear.next != null && count < k) {
count++;
rear = rear.next;
}
while (rear.next != null) {
rear = rear.next;
pre = pre.next;
}
System.out.println(pre.val);
}
}
}
3、完整题目
描述
输入一个单向链表,输出该链表中倒数第k个结点,链表的倒数第1个结点为链表的尾指针。
链表结点定义如下:
struct ListNode { int m_nKey; ListNode* m_pNext; };
正常返回倒数第k个结点指针,异常返回空指针.
要求:
(1)正序构建链表;
(2)构建后要忘记链表长度。
数据范围:链表长度满足 1≤n≤1000 , k≤n ,链表中数据满足 0≤val≤10000
本题有多组样例输入。
输入描述:
输入说明
1 输入链表结点个数
2 输入链表的值
3 输入k的值
输出描述:
输出一个整数
示例1
输入:
8 1 2 3 4 5 6 7 8 4
输出:
5