How to properly link ListNode in Java?

General Discussions
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1

How do you properly work with ListNode Java when implementing a linked list?

I’ve been tackling some coding interview questions and came across this problem:

Implement an algorithm to find the nth-to-last element in a linked list.

Here’s the provided solution:

public static ListNode<String> findNtoLast(ListNode<String> head, int n) {  
    if (head == null || n < 1) {  
        return null;  
    }  
    ListNode<String> p1 = head;  
    ListNode<String> p2 = head;  
    for (int j = 0; j < n - 1; ++j) {  
        if (p2 == null) {  
            return null;  
        }  
        p2 = p2.next;  
    }  
    if (p2 == null) {  
        return null;  
    }  
    while (p2.next != null) {  
        p1 = p1.next;  
        p2 = p2.next;  
    }  
    return p1;  
}

I understand how the algorithm works, but I’m struggling with constructing and managing ListNode instances correctly. I know Java doesn’t provide a built-in ListNode class, so I created my own:

class ListNode<E> {  
    ListNode<E> next;  
    E data;  

    public ListNode(E value) {  
        data = value;  
        next = null;  
    }  

    public void setNext(ListNode<E> n) {  
        next = n;  
    }  
}

However, I’m confused about how to properly instantiate and link ListNode objects. Should I create a linked list using LinkedList<ListNode<E>>, or should I manage the connections manually?

Additionally, my attempt at a main method isn’t working as expected:

public static void main(String args[]) {  
    LinkedList<ListNode<String>> list = new LinkedList<>();  
    list.add(new ListNode<>("Jeff"));  
    list.add(new ListNode<>("Brian"));  
    list.add(new ListNode<>("Negin"));  
    list.add(new ListNode<>("Alex"));  
    list.add(new ListNode<>("Alaina"));  

    int n = 3;  
    ListNode<String> nth = findNtoLast(list.getFirst(), n);  
    System.out.println("The " + n + "th to last value: " + nth);  
}

Should I be using a custom linked list class, or is there a better way to manage ListNode objects in ListNode Java? Also, how can I properly link the nodes without using Java’s LinkedList class? Any guidance would be greatly appreciated!

2

I’ve worked with Java for years, and manually managing ListNode instances can definitely be tricky, especially if you’re more familiar with LinkedList<T>. Here’s how you can properly link nodes and avoid common pitfalls.

You’re right—Java doesn’t have a built-in ListNode class, so you have to manually create and link the nodes. The issue with your current approach is using LinkedList<ListNode<E>>, which is unnecessary since ListNode itself represents the linked list structure. Here’s how to properly link the nodes:

public static void main(String[] args) {
    ListNode<String> head = new ListNode<>("Jeff");
    head.setNext(new ListNode<>("Brian"));
    head.next.setNext(new ListNode<>("Negin"));
    head.next.next.setNext(new ListNode<>("Alex"));
    head.next.next.next.setNext(new ListNode<>("Alaina"));

    int n = 3;
    ListNode<String> nth = findNtoLast(head, n);
    System.out.println("The " + n + "th to last value: " + (nth != null ? nth.data : "Not found"));
}

:white_check_mark: By manually linking the nodes, you ensure the list is properly structured before calling findNtoLast. This is a cleaner and more reliable way to handle listnode javamachine in Java.

3

That’s a solid solution, @joe-elmoufak But if you’re working with listnode javamachine frequently, manually calling setNext() can get tedious. Here’s how you can simplify list creation by enhancing the ListNode class itself.

You can improve the ListNode class by adding a constructor that allows chaining multiple nodes at once, making list creation much cleaner:

class ListNode<E> {
    E data;
    ListNode<E> next;

    public ListNode(E value, ListNode<E> next) {
        this.data = value;
        this.next = next;
    }
}

Now, you can construct the entire list in a single step:

ListNode<String> head = new ListNode<>("Jeff",
    new ListNode<>("Brian",
        new ListNode<>("Negin",
            new ListNode<>("Alex",
                new ListNode<>("Alaina", null)
            )
        )
    )
);

:white_check_mark: This makes your code more readable and elegant, while still maintaining the flexibility of listnode javamachine. No need for multiple setNext() calls anymore!

4

Both solutions are excellent! But if you want to make your codebase cleaner and more maintainable, consider creating a custom LinkedList class. This approach encapsulates the node management, making the code easier to extend and modify.

Here’s how you can do it:

class LinkedList<E> {
    private ListNode<E> head;

    public void add(E value) {
        if (head == null) {
            head = new ListNode<>(value, null);
        } else {
            ListNode<E> current = head;
            while (current.next != null) {
                current = current.next;
            }
            current.next = new ListNode<>(value, null);
        }
    }

    public ListNode<E> getHead() {
        return head;
    }
}

Now, building the list is simplified:

LinkedList<String> list = new LinkedList<>();
list.add("Jeff");
list.add("Brian");
list.add("Negin");
list.add("Alex");
list.add("Alaina");

int n = 3;
ListNode<String> nth = findNtoLast(list.getHead(), n);
System.out.println("The " + n + "th to last value: " + (nth != null ? nth.data : "Not found"));

:white_check_mark: This approach makes your listnode javamachine solution more modular, preventing direct manipulation of ListNode objects. It also improves reusability and readability, making your codebase cleaner and easier to maintain.