9.4. Linked Lists¶
9.4.1. Linked Lists¶
In this module we present one of the two traditional implementations for lists, usually called a linked list. The linked list uses dynamic memory allocation, that is, it allocates memory for new list elements as needed. The following diagram illustrates the linked list concept. Here there are three nodes that are "linked" together. Each node has two boxes. The box on the right holds a link to the next node in the list. Notice that the rightmost node has a diagonal slash through its link box, signifying that there is no link coming out of this box.
Because a list node is a distinct object (as opposed to simply a cell
in an array), it is good practice to make a separate list node class.
(We can also re-use the list node class to implement linked
implementations for the stack and
queue data structures.
Here is an implementation for list nodes, called the Link class.
Objects in the Link class contain an element field to
store the element value, and a next field to store a pointer to
the next node on the list.
The list built from such nodes is called a singly linked list,
or a one-way list, because each list node
has a single pointer to the next node on the list.
class Link { // Singly linked list node class
private Object e; // Value for this node
private Link n; // Point to next node in list
// Constructors
Link(Object it, Link inn) { e = it; n = inn; }
Link(Link inn) { e = null; n = inn; }
Object element() { return e; } // Return the value
Object setElement(Object it) { return e = it; } // Set element value
Link next() { return n; } // Return next link
Link setNext(Link inn) { return n = inn; } // Set next link
}
class Link { // Singly linked list node class
private Object e; // Value for this node
private Link n; // Point to next node in list
// Constructors
Link(Object it, Link inn) { e = it; n = inn; }
Link(Link inn) { e = null; n = inn; }
Object element() { return e; } // Return the value
Object setElement(Object it) { return e = it; } // Set element value
Link next() { return n; } // Return next link
Link setNext(Link inn) { return n = inn; } // Set next link
}
class Link<E> { // Singly linked list node class
private E e; // Value for this node
private Link<E> n; // Point to next node in list
// Constructors
Link(E it, Link<E> inn) { e = it; n = inn; }
Link(Link<E> inn) { e = null; n = inn; }
E element() { return e; } // Return the value
E setElement(E it) { return e = it; } // Set element value
Link<E> next() { return n; } // Return next link
Link<E> setNext(Link<E> inn) { return n = inn; } // Set next link
}
The Link class is quite simple.
There are two forms for its constructor, one with
an initial element value and one without.
Member functions allow the link user to get or set the element
and link fields.
9.4.1.1. Why This Has Problems¶
There are a number of problems with the representation just
described.
First, there are lots of special cases to code for.
For example, when the list is empty we have
no element for head, tail, and curr to point to.
Implementing special cases for insert and remove
increases code complexity, making it harder to understand,
and thus increases the chance of introducing bugs.
9.4.1.2. A Better Solution¶
Fortunately, there is a fairly easy way to deal with all of the special cases, as well as the problem with deleting the last node. Many special cases can be eliminated by implementing linked lists with an additional header node as the first node of the list. This header node is a link node like any other, but its value is ignored and it is not considered to be an actual element of the list. The header node saves coding effort because we no longer need to consider special cases for empty lists or when the current position is at one end of the list. The cost of this simplification is the space for the header node. However, there are space savings due to smaller code size, because statements to handle the special cases are omitted. We get rid of the remaining special cases related to being at the end of the list by adding a "trailer" node that also never stores a value.
The following diagram shows initial conditions for a linked list with header and trailer nodes.
Here is what a list with some elements looks like with the header and trailer nodes added.
Adding the trailer node also solves our problem with deleting the last node on the list, as we will see when we take a closer look at the remove method's implementation.
9.4.2. Linked List Implementation¶
Here is the implementation for the linked list class,
named LList.
// Linked list implementation
class LList implements List {
private Link head; // Pointer to list header
private Link tail; // Pointer to last element
private Link curr; // Access to current element
private int listSize; // Size of list
// Constructors
LList(int size) { this(); } // Constructor -- Ignore size
LList() { clear(); }
// Remove all elements
public void clear() {
curr = tail = new Link(null); // Create trailer
head = new Link(tail); // Create header
listSize = 0;
}
// Insert "it" at current position
public boolean insert(Object it) {
curr.setNext(new Link(curr.element(), curr.next()));
curr.setElement(it);
if (tail == curr) tail = curr.next(); // New tail
listSize++;
return true;
}
// Append "it" to list
public boolean append(Object it) {
tail.setNext(new Link(null));
tail.setElement(it);
tail = tail.next();
listSize++;
return true;
}
// Remove and return current element
public Object remove () {
if (curr == tail) return null; // Nothing to remove
Object it = curr.element(); // Remember value
curr.setElement(curr.next().element()); // Pull forward the next element
if (curr.next() == tail) tail = curr; // Removed last, move tail
curr.setNext(curr.next().next()); // Point around unneeded link
listSize--; // Decrement element count
return it; // Return value
}
public void moveToStart() { curr = head.next(); } // Set curr at list start
public void moveToEnd() { curr = tail; } // Set curr at list end
// Move curr one step left; no change if now at front
public void prev() {
if (head.next() == curr) return; // No previous element
Link temp = head;
// March down list until we find the previous element
while (temp.next() != curr) temp = temp.next();
curr = temp;
}
// Move curr one step right; no change if now at end
public void next() { if (curr != tail) curr = curr.next(); }
public int length() { return listSize; } // Return list length
// Return the position of the current element
public int currPos() {
Link temp = head.next();
int i;
for (i=0; curr != temp; i++)
temp = temp.next();
return i;
}
// Move down list to "pos" position
public boolean moveToPos(int pos) {
if ((pos < 0) || (pos > listSize)) return false;
curr = head.next();
for(int i=0; i<pos; i++) curr = curr.next();
return true;
}
// Return true if current position is at end of the list
public boolean isAtEnd() { return curr == tail; }
// Return current element value. Note that null gets returned if curr is at the tail
public Object getValue() {
return curr.element();
}
}
// Linked list implementation
class LList implements List {
private Link head; // Pointer to list header
private Link tail; // Pointer to last element
private Link curr; // Access to current element
private int listSize; // Size of list
// Constructors
LList(int size) { this(); } // Constructor -- Ignore size
LList() { clear(); }
// Remove all elements
void clear() {
curr = tail = new Link(null); // Create trailer
head = new Link(tail); // Create header
listSize = 0;
}
// Insert "it" at current position
boolean insert(Object it) {
curr.setNext(new Link(curr.element(), curr.next()));
curr.setElement(it);
if (tail == curr) tail = curr.next(); // New tail
listSize++;
return true;
}
// Append "it" to list
boolean append(Object it) {
tail.setNext(new Link(null));
tail.setElement(it);
tail = tail.next();
listSize++;
return true;
}
// Remove and return current element
Object remove () {
if (curr == tail) return null; // Nothing to remove
Object it = curr.element(); // Remember value
curr.setElement(curr.next().element()); // Pull forward the next element
if (curr.next() == tail) tail = curr; // Removed last, move tail
curr.setNext(curr.next().next()); // Point around unneeded link
listSize--; // Decrement element count
return it; // Return value
}
void moveToStart() { curr = head.next(); } // Set curr at list start
void moveToEnd() { curr = tail; } // Set curr at list end
// Move curr one step left; no change if now at front
void prev() {
if (head.next() == curr) return; // No previous element
Link temp = head;
// March down list until we find the previous element
while (temp.next() != curr) temp = temp.next();
curr = temp;
}
// Move curr one step right; no change if now at end
void next() { if (curr != tail) curr = curr.next(); }
int length() { return listSize; } // Return list length
// Return the position of the current element
int currPos() {
Link temp = head.next();
int i;
for (i=0; curr != temp; i++)
temp = temp.next();
return i;
}
// Move down list to "pos" position
boolean moveToPos(int pos) {
if ((pos < 0) || (pos > listSize)) return false;
curr = head.next();
for(int i=0; i<pos; i++) curr = curr.next();
return true;
}
// Return true if current position is at end of the list
boolean isAtEnd() { return curr == tail; }
// Return current element value. Note that null gets returned if curr is at the tail
Object getValue() {
return curr.element();
}
}
// Linked list implementation
class LList<E> implements List<E> {
private Link<E> head; // Pointer to list header
private Link<E> tail; // Pointer to last element
private Link<E> curr; // Access to current element
private int listSize; // Size of list
// Constructors
LList(int size) { this(); } // Constructor -- Ignore size
LList() { clear(); }
// Remove all elements
public void clear() {
curr = tail = new Link<E>(null); // Create trailer
head = new Link<E>(tail); // Create header
listSize = 0;
}
// Insert "it" at current position
public boolean insert(E it) {
curr.setNext(new Link<E>(curr.element(), curr.next()));
curr.setElement(it);
if (tail == curr) tail = curr.next(); // New tail
listSize++;
return true;
}
// Append "it" to list
public boolean append(E it) {
tail.setNext(new Link<E>(null));
tail.setElement(it);
tail = tail.next();
listSize++;
return true;
}
// Remove and return current element
public E remove () {
if (curr == tail) return null; // Nothing to remove
E it = curr.element(); // Remember value
curr.setElement(curr.next().element()); // Pull forward the next element
if (curr.next() == tail) tail = curr; // Removed last, move tail
curr.setNext(curr.next().next()); // Point around unneeded link
listSize--; // Decrement element count
return it; // Return value
}
public void moveToStart() { curr = head.next(); } // Set curr at list start
public void moveToEnd() { curr = tail; } // Set curr at list end
// Move curr one step left; no change if now at front
public void prev() {
if (head.next() == curr) return; // No previous element
Link<E> temp = head;
// March down list until we find the previous element
while (temp.next() != curr) temp = temp.next();
curr = temp;
}
// Move curr one step right; no change if now at end
public void next() { if (curr != tail) curr = curr.next(); }
public int length() { return listSize; } // Return list length
// Return the position of the current element
public int currPos() {
Link<E> temp = head.next();
int i;
for (i=0; curr != temp; i++)
temp = temp.next();
return i;
}
// Move down list to "pos" position
public boolean moveToPos(int pos) {
if ((pos < 0) || (pos > listSize)) return false;
curr = head.next();
for(int i=0; i<pos; i++) curr = curr.next();
return true;
}
// Return true if current position is at end of the list
public boolean isAtEnd() { return curr == tail; }
// Return current element value. Note that null gets returned if curr is at the tail
public E getValue() {
return curr.element();
}
}
Here are some special cases for linked list insertion: Inserting at the end, and inserting to an empty list.
