从LinkedList递归查找和删除节点

Question

Given a string to search for, I want to write a recursive function that takes in only one parameter (The string to search for). The function will search for the value recursively and if it is found then it will remove the item and return it. If it is not found then, the function will reach the end of the list and return null. What I have so far I think is the right idea except it is not functioning properly:

Main Test Class

public static void main(String[] args) {
    RecLinkedList list = new RecLinkedList();
    list.add("A");
    list.add("B");
    list.add("D");
    list.add("C", 2);
    list.add("E", 4);
    list.add("G", 6); //this should be invalid

    System.out.println( list );
    System.out.println( list.remove( 1 ).getValue() );
    System.out.println( list.remove("D").getValue() );
    System.out.println( list.remove("G").getValue() );
    System.out.println( list.size() );
    System.out.println( list );
}

Linked List Class (Showing only what I need help on)

public class RecLinkedList {
private Node first;
private int size = 0;

public RecLinkedList(){
    first = null;
}
public boolean isEmpty() {
    return first == null;
}
public Node remove( String s ){
    return remove( s, 0, first );
}
private Node remove( String s, int count, Node list ){
    if( list == null ){
        return null;
    }else if( s.equals(s) ){
        first = list.getNext();
        return list;
    }else if( s.equals(count+1) ){
        Node n = list.getNext();
        if( list.getNext() != null ){
            list.setNext( list.getNext().getNext() );
        }
        return n;
    }else{
        return remove( s, count+1, list.getNext() );
    }
}

So far, I am able to remove the item but as of now the item "A" is getting removed when it should not be. The final list should be A,C,E. (G should return and print null because it does not exist). I think I am close, but off by something minor, but I can not seem to figure it out.

Answer 1

There are several errors in your code (see comments below) :

private Node remove( String s, int count, Node list ){
    if( list == null ){
        return null;
    }else if( s.equals(s) ){ // comparing s to itself means you always remove
                             // the first element from the list (since this
                             // condition is always true)
        first = list.getNext();
        return list;
    }else if( s.equals(count+1) ){ // comparing the String s to an int - makes
                                   // no sense, will never be true
        Node n = list.getNext();
        if( list.getNext() != null ){
            list.setNext( list.getNext().getNext() );
        }
        return n;
    }else{
        return remove( s, count+1, list.getNext() );
    }
}

Answer 2

It seems to me like there is some ambiguity in your question. I understand that your method should search for an element, remove it if present, and return the same object. If the element is not present, the method should return null. That seems pretty straight-forward since most of your utility methods are already implemented in LinkedList . I thus recommend to extend that class:

public class RecLinkedList<E>
    extends LinkedList<E>
{
    public E removeAndReturn(E element)
    {
        E result;
        if (this.contains(element)) {
            remove(element);
            result = element;
        }
        else {
            result = null;
        }
        return result;
    }
}

I don't see why you would want to implement this recursively.

This could clearly be written more concisely, but the explicit if-else should make it clearer.

EDIT : The more concise and probably better implementation would be:

public E removeAndReturn(E element)
{
    return remove(element) ? element : null;
}