18. Delete the penultimate node of the linked list

Given a linked list , Delete the last of the linked list n A node and returns the header pointer of the linked list
for example ,
The list given is : 1\to 2\to 3\to 4\to 51→2→3→4→5, n= 2n=2.
Deleted the penultimate of the linked list nn After nodes , Linked list becomes 1\to 2\to 3\to 51→2→3→5.

Data range ： Chain length 0\le n \le 10000≤n≤1000, The value of any node in the linked list satisfies 0 \le val \le 1000≤val≤100
requirement ： Spatial complexity O(1)O(1), Time complexity O(n)O(n)
remarks ：
Title assurance nn It must be effective
Example 1
Input ：
{1,2},2    
Copy
Return value ：
{2} 

The code is as follows ：

import lombok.ToString;

@ToString
public class ListNode {
    int val;
    ListNode next = null;

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

public class Main {
    public static void main(String[] args) throws Exception {
        ListNode head = new ListNode(1);
        ListNode headOne = new ListNode(2);
        ListNode headTwo = new ListNode(3);
        head.next = headOne;
        headOne.next = headTwo;
        System.out.println(head);
        System.out.println(removeNthFromEnd(head, 3));
    }

    public static ListNode removeNthFromEnd (ListNode head, int n) {
        ListNode first = head;
        ListNode second = head;
        for(int i = 0; i < n; i++) {
            first = first.next;
        }
        // If n The value of is equal to the length of the linked list , Directly return to the linked list of U-turn nodes 
        if (first == null) {
            return head.next;
        }
        while(first.next != null) { // Move both pointers at the same time 
            first = first.next;
            second = second.next;
        }
        second.next = second.next.next;
        return head;
    }
}