CODEWITHSUNDEEP
javalinked-listnodes
damon
damon

Reputation: 8467

removing node in linkedlist using java

I am trying to code a method for deleting the last Node in a linkedlist(for learning how to modify a linkedlist..I am not using the java library LinkedList class)..I tried to deal with the usecase where the passed linkedlist has only one Node.

However when I try to print the linkedlist before and after the removal, it gives the same output as if the node was not removed.

class NodeProcessing{
    public static void removeLastNode(Node f){
        if (f==null) return;
        if(f.next == null){//if linkedlist has single node
            f = null;
            return;
        }
        ...
    }

    public static void showList(Node first){
        System.out.println("linked list=");
        for(Node x = first; x != null; x = x.next){
            System.out.print(x.item+" ,");
        }
        System.out.println();
    }

    public static void main(String[] args) {
        Node a = new Node();
        a.item = "one";
        showList(a);
        removeLastNode(a);
        showList(a);
    }

}
class Node{
    String item;
    Node next;
}

Output:

linked list= one ,

linked list= one ,

update: when I used the debugger,I can see that the Node a in main() has address :Node@a61164 and the Node f inside removeLastNode() also has :Node@a61164

Upvotes: 1

Views: 3394

Answers (3)

codeMan
codeMan

Reputation: 5758

The Node f is local variable in the following method

public static void removeLastNode(Node f)

Since it is a copy of the reference to the actual argument, the changes made to the parameters will not affect the arguments.

you are doing

f = null

where as

Node first

will be still pointing to the existing node. You are using the argument to print the linked list, Hence the result.

Upvotes: 1

Prasad Kharkar
Prasad Kharkar

Reputation: 13556

You can use the java provided LinkedList data structure to do this.

    LinkedList items = new LinkedList();
    items.add("one");
    items.add("two");

    System.out.println(items);
    items.removeLast();
    System.out.println(items);

This will generate the desired output.

Upvotes: 0

Jon Skeet
Jon Skeet

Reputation: 1499770

Setting f to null does nothing - that just changes the value of the parameter, which is just a local variable. It's important to understand that Java always uses pass-by-value for parameters - the value is a reference when the parameter type is a class type, but that reference is passed by value. Changing the parameter value to a different reference doesn't change the caller's argument.

Ideally you'd want to separate the concept of "the list" from "a node in the list" (just as the Java API does). That way when you call remove on a list, it mutates the list itself. In your current model of "a node is just the head of a list" there's no way you can remove the final node - you can't destroy the node itself.

The closest you could come would be to make removeLastNode return a Node reference - which could return null if the notional list is now empty.

Upvotes: 6

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