将指向ArrayList的LinkedList的指针转换为String

Question

I'm trying to make a linkedList (elements) of custom arrayList (coordinated )that with type Object the Output I am looking for is

Arraylist1 coord1 [2,5,1]
Arraylist2 coord2 [7,6,9]
LinkedList List1 [[2,5,1],[7,6,9]]

this is my output

[ 2 5 1 ] [ 7 6 9 ]

assignment1.arrayList@33909752

assignment1.arrayList@55f96302

I tried using toString method with no luck can someone please explain to me how to print the output with no pointers to memory. and do I need a method to access a specific element in a certain position (in the linkList) and get its coordinates?

here is my code: thanks

public class arrayList {
private Object[] myList;
private int counter = 0;
private int capacity = 100;

public arrayList() {
    myList = new Object[this.capacity];
}

public Object get(int index) {
    if (index < counter) {
        return myList[index];
    } else {
        throw new ArrayIndexOutOfBoundsException();
    }
}

public void add(Object obj) {
    myList[counter++] = obj;
}

public Object remove(int index) {
    if (index < counter) {
        Object obj = myList[index];
        int temp = index;
        myList[index] = null;

        while (temp < counter) {
            myList[temp] = myList[temp + 1];
            myList[temp + 1] = null;
            temp++;
        }

        counter--;
        return obj;
    } else {
        throw new ArrayIndexOutOfBoundsException();
    }
}

public int size() {
    return counter;
}

public void display(Object obj1) {
    System.out.print("[");
    for (int i = 0; i < this.size(); i++) {
        System.out.print(" " + this.get(i) + " ");
    }
    System.out.print("]");
    System.out.println();

}

}

public class linkedList {

public Cube firstLink;
public Cube next;

linkedList() {
    firstLink = null;
}

public void insertFirstLink(Object e) {

    Cube newLink = new Cube(e);

    newLink.next = firstLink;
    firstLink = newLink;
}

public boolean isEmpty() {
    return (firstLink == null);
}

public Cube removeFirst() {

    Cube linkReference = firstLink;

    if (!isEmpty()) {
        firstLink = firstLink.next;
    } else {
        System.out.println("Empty Linked list!");

    }
    return linkReference;

}

public void display() {

    Cube theLink = firstLink;

    while (theLink != null) {
        theLink.display();
        theLink = theLink.next;
        System.out.println();
    }
}

public Cube find(Object obj) {
    Cube theLink = firstLink;
    if (!isEmpty()) {
        while (theLink.obj != obj) {
            if (theLink.next == null) {
                return null;
            } else {
                theLink = theLink.next;
            }
        }

    } else {
        System.out.println("Empty List!");
    }

    return theLink;
}

public Cube removeLink(Object obj) {
    Cube currentLink = firstLink;
    Cube previousLink = firstLink;

    while (currentLink.obj != obj) {
        if (currentLink.next == null) {
            return null;
        } else {
            previousLink = currentLink;
            currentLink = currentLink.next;
        }
    }

    if (currentLink == firstLink) {
        firstLink = firstLink.next;
    } else {

        previousLink.next = currentLink.next;
    }
    return currentLink;

}

}

public class Cube {
public Object obj;
public Cube next;

public Cube(Object obj) {
    this.obj = obj;

}

public void display() {
    obj.toString();
    System.out.println(obj);
}

public static void main(String[] args) {

    arrayList coord1 = new arrayList();
    coord1.add(new Integer(2));
    coord1.add(new Integer(5));
    coord1.add(new Integer(1));

    arrayList coord2 = new arrayList();
    coord2.add(new Integer(7));
    coord2.add(new Integer(6));
    coord2.add(new Integer(9));


    coord1.display(coord1);
    coord2.display(coord2);


    linkedList position1 = new linkedList();
    position1.insertFirstLink(coord1);
    position1.insertFirstLink(coord2);

    position1.display();


}

}

Answer 1

According to the API, https://docs.oracle.com/javase/7/docs/api/java/io/PrintStream.html#println(java.lang.Object)

You put a Object into System.out.println(x) will eventually call String.valueOf(x) which utlimitly call x.toString() . Put this #1

In linkedList.display() it will call Cube.display() . Cube.display() will prints its encapsulated object. The object turns out to be an arrayList . From #1, the method call would become arrayList.toString() . However, arrayList did not implement toString() . Don't worry, Java got your back. Since all non-primitive type extends Object, arrayList will has is toString() method inherited from Object. This is the Object implementation, to print the object id. If you don't like it, you can define your own toString() in arrayList scope.

Update 1

You may wonder why the first arraylist could print something meaningful out. It is because you call its arraylist.display() directly, not System.out.println(coord1)

Update 2

There are some anti pattern in your code.

1. Class name should be in camel case , like Cube so as ArrayList, LinkedList.
2. Why arrayList.display() need a argument which also reference to itself but you never use it (ie obj1 is totally ignored)? coord1.display() is good enough, you don't need coord1.display(coord1)

Answer 2

You need to do following changes:

In your arrayList class you need to change methods display(Object obj1) and override toString() method:

public void display(Object obj1) {
    System.out.println(obj1);
}

@Override
public String toString() {

    StringBuilder sb = new StringBuilder();
    sb.append("[");
    for (int i = 0; i < counter - 1; i++) {
        sb.append(myList[i] + ",");
    }

    sb.append(myList[counter - 1] + "]");
    return sb.toString();
}

In class linkedList change display and override toString() as below:

public void display() {
    System.out.println(this.toString());
}

@Override
public String toString() {
    StringBuilder sb = new StringBuilder();
    Cube temp = firstLink;
    sb.append("[");
    while (temp != null) {
        sb.append(temp.obj.toString());
        temp = temp.next;
        if (temp != null)
            sb.append(",");
    }
    sb.append("]");

    return sb.toString();
}

}

Note: It will be printing linked list in reverse order, because your implementation of LinkedList is storing data in reverse order.