LinkedList Get Method

Question

I have a get method for my only linked list and they work fine, but my instructor told me that he wants me to shorten the code because I have too many special cases. The problem is that when I try to cut some parts of my code, the code no longer works as intended.

Code written from me:

public E get(int index) {
    Node<E> f = first;

    // If index is bigger / smaller than Linked List size throw IndexOutOfBounds
    if (index > size || index < 0){
        throw new IndexOutOfBoundsException();
    }


    // Index Less than size and is not the first or last node.
    if (index < size && index > 0) {
        for (int i = 0; i < index; i++) {
            f = f.next;
        }
        return f.getValue();
    }

    // If the Linked List is empty + Index = 0 return null
    if (first == null && index == 0) {
        return null;
    }

    // If Linked List is not empty and index = 0 return first value
    if (index == 0) {
        return first.getValue();
    }

    // If index = end of list
    if (index == size) {
        return last.getValue();
    }


    // Return null if not found.
    return null;

}

Therefore, he tells me that I delve too much into this, and only two cases are needed if the index is valid or invalid; with which I agree with him, so I'm trying to shorten the code to this:

Node<E> f = first;

// If index is bigger / smaller than Linked List size throw IndexOutOfBounds
if (index > size || index < 0){
    throw new IndexOutOfBoundsException();
}


// Index Less than size and is not the first or last node.
if (index <= size && index >= 0) {
    for (int i = 0; i < index; i++) {
        f = f.next;
    }
}
return f.getValue();

and for my test cases, I use a Linked List with these values:

[Frank, George, Josh, Jim, Groom, Susie, John, Jim, Dakota, Levy, Jackson, Jeff, Walt, Matt]

and my test cases in my test class are as follows:

System.out.println("Get Method Test ----------------------------");
System.out.println("First Index: " + ll.get(0));
System.out.println("Last Index: " + ll.get(ll.size()));
System.out.println("'Middle' Index: " + ll.get(5));
System.out.println("Empty list with index of 0: " + llempty.get(0));

a NullPointerException :

:

First Index: Frank
Exception in thread "main" java.lang.NullPointerException

:

: , , "" , 0, ,

, /, !

Answer 1

. , , , n n-1. get():

// If index is bigger / smaller than Linked List size throw IndexOutOfBounds
if (index > size || index < 0){
    throw new IndexOutOfBoundsException();
}

index >= size :

// If index is bigger / smaller than Linked List size throw IndexOutOfBounds
if (index >= size || index < 0){
    throw new IndexOutOfBoundsException();
}

.

System.out.println("Last Index: " + ll.get(ll.size()));

ll.size() - 1 :

System.out.println("Last Index: " + ll.get(ll.size() - 1));

. , .

, IndexOutOfBoundsException, . "Index Too Small" "Index Too Big" IndexOutOfBoundsException.

Answer 2

( ), "", , 0 ( ),

index == size -1

Answer 3

( if ):

//assume 0 based index
int current = 0;

//loop until we hit the requested index
while (current != index) {
 f = f.next;
 current++;
}

//return the right node
return f;

Assuming your initial check of the index outside is acceptable, this should never cause a problem. However, if your indexes are based on 0, you need to change your check beyond:

if (index > size - 1 || index < 0) {

For example, if there are 2 elements, you have indexes 0 and 1. In this case, index 2 is not valid.

Answer 4

Must be

if (index >= size || index < 0){
        throw new IndexOutOfBoundsException();
    }

if (index < size && index >= 0) {
    for (int i = 0; i < index; i++) {
        f = f.next;
    }
}