Implementing Lists Using Linked-Lists

Contents

• Introduction
• Java Types
  • Test Yourself #1
• Intro to Linked Lists
  • Test Yourself #2
• Linked List Operations
  • Adding a node
    • Test Yourself #3
  • Removing a node
  • Using a header node
• The LinkedList Class
  • add (to end of list)
    • Test Yourself #4
  • add (at a given position)
    • Test Yourself #5
  • The LinkedList constructor
    • Test Yourself #6
• Comparison: Lists via Arrays versus via Linked Lists
  • Test Yourself #7
• Linked List Variations
  • Doubly linked lists
  • Circular linked lists
    • Test Yourself #8
  • Comparisons

The previous set of notes discussed how to implement the ListADT interface using an array to store the items in the list. Here we discuss how to implement the ListADT interface using a linked list to store the items. However, before talking about linked lists, we will review the difference between primitive and non-primitive types in Java.

Java has two "categories" of types:

1. primitive types: short, int, long, float, double, boolean, char, and byte
2. reference types: arrays and classes

When you declare a variable with a primitive type, you get enough space to hold a value of that type. Here's some code involving a primitive type, and the corresponding conceptual picture:

When you declare a variable with a reference type, you get space for a reference (or pointer) to that type, not for the type itself. You must use "new" to get space for the type itself. This is illustrated below.

Remember that class objects are also reference types. For example, if you declare a variable of type List, you only get space for a pointer to a list; no actual list exists until you use "new". This is illustrated below, assuming the array implementation of lists, and assuming that the List constructor initializes the items array to be of size 3. Note that because it is an array of Objects, each array element is (automatically) initialized to null (shown using a diagonal line in the picture).

An important consequence of the fact that non-primitive types are really pointers is that assigning from one variable to another can cause aliasing (two different names refer to the same object). For example:

Note that in this example, the assignment to B[1] changed not only that value, but also the value in A[1] (because A and B were pointing to the same array)! However, an assignment to B itself (not to an element of the array pointed to by B) has no effect on A:

A similar situation arises when a non-primitive value is passed as an argument to a method. For example, consider the following 4 statements, and the definition of method changeArray:

1.   int[] A = new int[3];
2.   A[1] = 6;
3.   changeArray(A);
4.   System.out.print(A[1]);

5.   public static void changeArray(int[] X) {
6.     X[1] = 10;
7.     X = new int[2];
8.   }

The picture below illustrates what happens when this code executes.

Note that the method call causes A and X to be aliases (they both point to the same array). Therefore, the assignment X[1] = 10 changes both X[1] and A[1]. However, the assignment X = new int[2] only changes X, not A, so when the call to changeArray finishes, the value of A[1] is still 10, not 0.

For each line of the code shown below, draw the corresponding conceptual picture.

int [] X = new int[2];
X[0] = 1;
int Y = new int[3];
Y[0] = 2;
Y = X;
Y[0] = 3;

solution

Note that a linked list consists of one or more nodes. Each node contains some data (in this example, item 1, item 2, etc,) and a pointer. For each node other than the last one, the pointer points to the next node in the list. For the last node, the pointer is null (indicated in the example using a diagonal line). To implement linked lists in Java, we will define a Listnode class, to be used to represent the individual nodes of the list.

public class Listnode {
  //*** fields ***
    private Object data;
    private Listnode next;

  //*** methods ***
    // 2 constructors
    public Listnode(Object d) {
	this(d, null);
    }

    public Listnode(Object d, Listnode n) {
	data = d;
	next = n;
    }
    
    // access to fields
    public Object getData() {
        return data;
    }

    public Listnode getNext() {
        return next;
    }

    // modify fields
    public void setData(Object ob) {
        data = ob;
    }

    public void setNext(Listnode n) {
        next = n;
    }
}

To understand this better, consider writing code to create a linked list with two nodes, containing "ant" and "bat", respectively, pointed to by a variable named L. First we need to declare variable L; here's the declaration together with a picture showing what we have so far:

To make L point to the first node of the list, we need to use new to allocate space for that node. We want its data field to contain "ant" and (for now) we don't care about its next field, so we'll use the 1-argument Listnode constructor (which sets the next field to null):

To add the second node to the end of the list we need to create the new node (with "bat" in its data field and null in its next field), and we need to set the next field of the first node to point to the new one:

Assume that the list shown above (with nodes "ant" and "bat") has been created.

Question 1.
Write code to change the contents of the second node's data field from "bat" to "cat".

Question 2.
Write code to insert a new node with "rat" in its data field between the two existing nodes.

solution

• Adding a node after a given node in the list.
• Removing a given node from the list.

1. n, (a pointer to) a node in a list (i.e., n is a Listnode), and
2. newdat, the data to be stored in a new node

Step 1: create the new node using the given data

Step 2: "link it in":
(a) make the new node's next field point to whatever n's next field was pointing to
(b) make n's next field point to the new node.

And here's the code:

Listnode tmp = new Listnode(newdat);    // Step 1
tmp.setNext( n.getNext() );             // Step 2(a)
n.setNext( tmp );                       // Step 2(b)

TEST YOURSELF #3

Draw pictures like the ones given above, to illustrate what happens when node n is the last node in the list. Does the statement

n.setNext( new Listnode(newdat, n.getNext()) );

solution

Now consider the worst-case running time for this add operation. Whether we use the single statement or the list of three statements, we are really doing the same thing:

1. Using new to allocate space for a new node (start step 1).
2. Initializing the new node's data and next fields (finish step 1 + step 2(a)).
3. Changing the value of n's next field (step 2(b)).

Note that the fact that n's next field is still pointing to a node in the list doesn't matter -- n has been removed from the list, because it cannot be reached from L. It should be clear that in order to implement the remove operation, we first need to have a pointer to the node before node n (because that node's next field has to be changed). The only way to get to that node is to start at the beginning of the list. We want to keep moving along the list as long as the current node's next field is not pointing to node n. Here's the appropriate code:

Listnode tmp = L;
while (tmp.getNext() != n) tmp = tmp.getNext();  // find the node before n

Note that this kind of code (moving along a list until some condition holds) is very common. For example, similar code would be used to implement a lookup operation on a linked list (an operation that determines whether there is a node in the list that contains a given piece of data).

Note also that there is one case when the code given above will not work. When n is the very first node in the list, the picture is like this:

In this case, the test (tmp.getNext() != n) will always be false, and eventually we will "fall off the end" of the list (i.e., tmp will become null, and we will get a runtime error when we try to dereference a null pointer). We will take care of that case in a minute; first, assuming that n is not the first node in the list, here's the code that removes n from the list:

Listnode tmp = L;
while (tmp.getNext() != n) tmp = tmp.getNext();  // find the node before n 
tmp.setNext( n.getNext() );                      // remove n from the linked list

How can we test whether n is the first node in the list, and what should we do in that case? If n is the first node, then L will be pointing to it, so we can test whether L == n. The following before and after pictures illustrate removing node n when it is the first node in the list:

Here's the complete code for removing node n from a linked list, including the special case when n is the first node in the list:

if (L == n) {
  // special case: n is the first node in the list
    L = n.getNext();
} else {
  // general case: find the node before n, then "unlink" n
    Listnode tmp = L;
    while (tmp.getNext() != n) tmp = tmp.getNext();
    tmp.setNext( n.getNext() );
}

What is the worst-case running time for this remove operation? If node n is the first node in the list, then we simply change one field (L's next field). However, in the general case, we must traverse the list to find the node before n, and in the worst case (when n is the last node in the list), this requires time proportional to the number of nodes in the list. Once the node before n is found, the remove operation involves just one assignment (to the next field of that node), which takes constant time. So the worst-case time running time for this operation on a list with N nodes is O(N).

Using a header node

The LinkedList Class

Look back at the definition of the ArrayList class in the second set of notes and think about which fields and/or methods need to be changed before reading any further.

Clearly, the type of the items field needs to change, since the items will no longer be stored in an array; instead, the items will be stored in a linked list. Since having a header node simplifies the add and remove operations, we will assume that the linked list has a header node, Here's new declaration for the items field:

    private Listnode items;  // pointer to the header node of the list of items

Given this declaration for the items field, let's think again about the three List methods that were discussed assuming the array implementation:

1. The version of add that adds to the end of the list.
2. The version of add that adds to a given position in the list.
3. The constructor.

add (to end of list)

The disadvantage of this approach is that it requires O(N) time to add a node to the end of a list with N items. An alternative is to add a lastNode field (often called a tail pointer) to the LinkedList class, and to implement the methods that modify the linked list so that lastNode always points to the last node in the linked list (which will be the header node if the list is empty). There is more opportunity for error (since several methods will need to ensure that the lastNode field is kept up to date), but the use of the lastNode field will mean that the worst-case running time for this version of add is always O(1), and that will be important for applications that frequently add to the end of a list (which is often a common operation).

Here's a picture of the "ant, bat, cat" list, when the implementation includes a lastNode pointer:

Write the "add at the end" method (assuming that the LinkedList class includes both a header node and a lastNode field).

solution

add (at a given position)

TEST YOURSELF #5

Write the "add at a given position" method (assuming that the LinkedList class includes both a header node and a lastNode field).

solution

The LinkedList constructor

1. Listnode items (the pointer to the header node)
2. Listnode lastNode (the pointer to the last node in the list)
3. int numItems

Write the constructor.

solution

• space requirements
• time requirements
• ease of implementation

• In the linked-list implementation, one pointer must be stored for every item in the list, while the array stores only the items themselves.
• On the other hand, the space used for a linked list is always proportional to the number of items in the list. This is not necessarily true for the array implementation as described: if a lot of items are added to a list and then removed, the size of the array can be arbitrarily greater than the number of items in the list. However, we could fix this problem by modifying the remove operation to shrink the array when it becomes too empty.

• Adding an item to the end of a list is O(1) for the array implementation if we can use a "shadow" array (as discussed in class) to avoid the O(N) cost of calling method expandArray. It is also O(1) for the linked-list implementation as long as we have a lastNode field.
• Adding an item at a given position requires O(N) worst-case time for the array implementation, because existing items need to be moved. The operation is also O(N) in the worst case for the linked-list implementation, because we have to find the node currently in that position. So this operation is worst-case O(N) for both implementations. However, it is worth noting that for the array implementation adding closer to the beginning of the list is worst, and adding toward the end of the list is best, while it is the other way around for the linked-list implementation.
• The get operation is O(1) for the array implementation, and worst-case O(N) for the linked-list implementation.

In terms of ease of implementation, straightforward implementations of both the array and linked-list versions seem reasonably easy. However, the methods for the linked-list version seem to require more special cases.

Assume that lists are implemented using linked lists. How much time is required (using Big-O notation) to remove the first item from a list? to remove the last item from a list? How do these times compare to the times required for the same operations when the list is implemented using an array?

solution

1. doubly linked lists
2. circular linked lists

Another way to fix the problem is to use a doubly linked list. Here's the conceptual picture:

Each node in a doubly linked list contains three fields: the data, and two pointers. One pointer points to the previous node in the list, and the other pointer points to the next node in the list. The previous pointer of the first node, and the next pointer of the last node are both null. Here's the Java class definition for a doubly linked list node:

public class DblListnode {
  //*** fields ***
    private DblListnode prev;
    private Object data;
    private DblListnode next;

  //*** methods ***
    // 3 constructors
    public DblListnode() {
	this(null, null, null);
    }

    public DblListnode(Object d) {
	this(null, d, null);
    }

    public DblListnode(DblListnode p, Object d, DblListnode n) {
        prev = p;
	data = d;
	next = n;
    }

    // access to fields
    public Object getData() {
        return data;
    }

    public DblListnode getNext() {
        return next;
    }

    public DblListnode getPrev() {
        return prev;
    }

    // modify fields
    public void setData(Object ob) {
        data = ob;
    }

    public void setNext(DblListnode n) {
        next = n;
    }

    public void setPrev(DblListnode p) {
        prev = p;
    }
}

To remove a given node n from a doubly linked list, we need to change the prev field of the node to its right, and we need to change the next field of the node to its left, as illustrated below.

Here's the code for removing node n:

// Step 1: change the prev field of the node after n
  DblListnode tmp = n.getNext();
  tmp.setPrev( n.getPrev() );

// Step 2: change the next field of the node before n
  tmp = n.getPrev();
  tmp.setNext( n.getNext() );

The class definitions are the same as for the non-circular versions. The difference is that, instead of being null, the next field of the last node points to the first node, and (for doubly linked circular lists) the prev field of the first node points to the last node.

The code given above for removing node n from a doubly linked list will work correctly except when node n is the first node in the list. In that case, the variable L that points to the first node in the list needs to be updated, so special-case code will always be needed unless the list includes a header node.

Another issue that you must address if you use a circular linked list is that if you're not careful, you may end up going round and round in circles! For example, what happens if you try to search for a particular value val using code like this:

Listnode tmp = L;
while (tmp != null && !tmp.getData().equals(val)) tmp = tmp.getNext();

TEST YOURSELF #8

Write the correct code to search for value val in a circular linked list pointed to by L. (Assume that the list contains no null data values.)

solution

The major advantage of doubly linked lists is that they make some operations (like the removal of a given node, or a right-to-left traversal of the list) more efficient.

The major advantage of circular lists (over non-circular lists) is that they eliminate some special-case code for some operations. Also, some applications lead naturally to circular list representations. For example, a computer network might best be modeled using a circular list.