Lecture06 methods for-making_data_structures_v2

Chapter 6Methods for Making

Data Structures

Dynamic Arrays in Data Structures

• In almost every data structure, we want functions for inserting and removing data.

• When dynamic arrays are used, the insertion function would add data to the array, while the removal function would “eliminate” data from the array (make it unusable).

• When the array becomes full, we would want to do an expansion – when many elements have been removed, we would want to do a contraction, so that only the used elements remain.

Array Expansion/Contraction

• One possible method:– When an element is inserted by the client,

increase the size of the array by 1.– When an element is removed by the client,

decrease the size of the array by 1.• The problem with this method is that it is

inefficient – every time an element is inserted or removed, the changeSize function is called…

changeSize Function

25 75 10 12 56 32 73 87… 0 1 2 3 432 433 444 445

New element needs to be put into array, so changeSize function is called

changeSize Function(cont.)

25 75 10 12 56 32 73 87… 0 1 2 3 432 433 444 445

… 0 1 2 3 432 433 444 445 446

new array is made

25 75 10 12 56 32 73 87… 0 1 2 3 432 433 444 445

elements are copied over one by one using a for loop

25 75 10 12 56 32 73 87… 0 1 2 3 432 433 444 445 446

Then, the new element can be put in

25 75 10 12 56 32 73 87… 0 1 2 3 432 433 444 445 446

This process would take place every time a new element needs to be inserted.

Suppose the element at the end of the array needs to be removed.

25 75 10 12 56 32 73 87… 0 1 2 3 432 433 444 445 446

Likewise, when an element needs to be removed, this method contracts the array by one to conserve memory.

The changeSize function is called and a new, smaller array is made.

25 75 10 12 56 32 73 87… 0 1 2 3 432 433 444 445 446

… 0 1 2 3 432 433 444 445

The elements are copied over one by one, using a for loop.

25 75 10 12 56 32 73 87… 0 1 2 3 432 433 444 445 446

25 75 10 12 56 32 73 87… 0 1 2 3 432 433 444 445

This method of array expansion/contraction is largely inefficient, because there is too much element copying.

25 75 10 12 56 32 73 87… 0 1 2 3 432 433 444 445

Linked Structures

• Sometimes it is best to store data in a linked structure (an alternative to an Array)

• A linked structure consists of a group of nodes – each node is made from a struct.

• An object of the Node struct contains an element of data.

A Node Struct Template

template <typename T>struct Node {

T info;Node<T> *next;

The info member is for the data. It can anything (T), but it is often the object of another struct, used as a record of information.

The next pointer stores the address of a Node of the same type! This means that each node can point to another node.

• In a data structure, each node is made in the heap; therefore, a node can only be accessed by a pointer.

• The client does not deal with nodes. • When the client uses an insertion function,

an element of data is passed into the insert function, and the function places it in a node.

Nodes (cont.)

• When the client wants to retrieve data, the data in a node is returned to the client (but not the node itself).

• The node struct template exists for use by the data structure.

Example of a Linked Structure

Each blue node is divided into two sections, for the two members of the Node struct.

Example of a Linked Structure (cont.)

The right section is the pointer called “next”.

The left section is the info member.

Example of a Linked Structure (cont.)

The last node doesn’t point to another node, so its pointer (called next) is set to NULL (indicated by slash).

The start pointer would be saved in the private section of a data structure class.

Linked Lists

• The arrangement of nodes in the linked structure on the previous slide is often called a linked list.

• We can access any element of the linked list, for retrieval of information.

• We can also remove any element from the linked list (which would shorten the list).

• We can also insert any element into any position in the linked list.

Linked ListAdvantages

… …5 3 7 2 1

Removing an element from the middle of a linked list is fast.

Linked ListAdvantages (cont.)

… …5 3 2 1

Removing an element from the middle of a linked list is fast.

Removal Problem in Array

… …

Removing elements from the middle of an array (without leaving gaps) is more problematic.

25 75 10 12

211 212 213 214 215 216 217 218

33 49 29 87

Removal Problem in Array (cont.)

… …

A loop must be used to slide each element on the right one slot to the left, one at a time…

25 75 10

211 212 213 214 215 216 217 218

33 49 29 87

… …25 75 10

211 212 213 214 215 216 217 218

49 29 8733

… …25 75 10

211 212 213 214 215 216 217 218

49 29 8733

… …25 75 10

211 212 213 214 215 216 217 218

49 29 8733

… …25 75 10

211 212 213 214 215 216 217 218

49 29 8733

Only 100,000 more to go!

• Linked lists also waste less memory for large elements (records of information).

• Wasted memory is memory space in the data structure not used for data.

• In arrays, the wasted memory is the part of the array not being utilized.

• In linked lists, the wasted memory is the pointer in each node.

Linked List

Accessing info

To access the info in the first node:

(*start).info

Or (better yet)

start->info

dereference and member access in one shot

Accessing info(cont.)

To access the info in the second node:

start->next->info

Finding a Possible Mercedes

Let’s solve the problem, but let’s assume that item is passed in as a parameter (of type T). This is normally what would happen.Instead of the CarType struct having an overloaded != operator, it will have an overloaded == operator.

itemmaker: Mercedes price: year:operator ==

CarType item;item.maker = "Mercedes";Node<T> *ptr = start;bool found = false;while (ptr != NULL && !found ) {

if ( ptr->info == item ) // overloaded ==found = true;

if ( !found )ptr = ptr->next;

Finding a Possible Mercedes (cont.)

CarType item;item.maker = "Mercedes";

Node<T> *ptr = start;bool found = false;while (ptr != NULL && !found ) {

CarType item;item.maker = "Mercedes";Node<T> *ptr = start;

bool found = false;while (ptr != NULL && !found ) {

found: false

CarType item;item.maker = "Mercedes";Node<T> *ptr = start;bool found = false;

while (ptr != NULL && !found ) {if ( ptr->info == item ) // overloaded ==

found = true;if ( !found )

ptr = ptr->next;}

found: false

if ( ptr->info == item ) found = true;

found: false

if ( !found )

ptr = ptr->next;}

found: false

while (ptr != NULL && !found ) { if ( ptr->info == item )

ptr = ptr->next;}

found: false

if ( !found ) ptr = ptr->next;

found: false

ptr = ptr->next;}

found: false

After going through the loop several times…

found: false

Notice that found is only set to true if ptr is not NULL and Mercedes is found …

ptr = ptr->next;}

found: false

then, !found is false and the loop exits

found: false

If Mercedes is not found, ptr eventually gets set to NULL.

What If Mercedes Does Not Exist?

if ( !found )

ptr = ptr->next;}

ptr is set to NULL

found: false

What If Mercedes Does not Exist? (cont.)

If Mercedes is not found, ptr eventually gets set to NULL.

ptr = ptr->next;}

ptr is set to NULL

found: false

What If Mercedes Does not Exist? (cont.)

Exit from loop because ptr is NULL.

What If Finding in an Empty Linked List?

• When a linked list is empty, the start pointer should always be set to NULL.

• The start pointer would be set to NULL inside the constructor, when an empty linked list is first made.

start is set to NULL

SAME CODE

Finding in an Empty List

ptr is set to NULL

Finding in an Empty List (cont.)

ptr is set to NULL

found: false

ptr is set to NULL

found: false

Exit loop because ptr is NULL.

Inserting a New Node

• Let’s assume that we want to insert a new node at the beginning of a linked list.

• Assume that the client passes in a parameter called element (of type T).

• We would like to place the element into a node and insert the node at the beginning of the linked list.

Inserting a Node at Front

element

All new nodes must be made in the heap, SO…

element

Node<T> *ptr = new Node<T>;

Inserting a Node at Front (cont.)

element

Node<T> *ptr = new Node<T>;

Now we have to store element into the node

element

Node<T> *ptr = new Node<T>;ptr->info = element;

element

ptrNow we have to think about how to make the pointer called “next” point to the first node in the list, to link it in

element

ptrYou can’t successfully write code like this without thinking about addresses.

element

ptrREMEMBER…when you want to change the way a pointer points, you HAVE to assign a different address to it

element

ptrRight now, the pointer called “next” doesn’t have a valid address assigned to it.

element

ptrTo store the correct address in it, we have to find the address of the first node of the linked list.

element

ptr Where is the address of the first node stored?

element

ptrNow think, the address would be stored in something that points to it. So where is it stored?

element

ptr That’s right, in the start pointer.

element

ptr So now, all we have to do is copy that address into the pointer called “next”

element

Node<T> *ptr = new Node<T>;ptr->info = element;ptr->next = start;

element

Well, it’s been inserted. But start should point to the first node now.

element

REMEMBER…when you want to change the way a pointer points, you have to assign a different address to it

element

We’d like start to point to the new node, so what stores the address of the new node?

element

That’s right, ptr. So now all we have to do is assign the address stored in ptr to the start pointer.

element

Node<T> *ptr = new Node<T>;ptr->info = element;ptr->next = start;start = ptr;

element

Node<T> *ptr = new Node<T>;ptr->info = element;ptr->next = start;start = ptr;

Easy, right?

REMEMBER…

• Use drawings when working with linked lists, until you become an expert.

• When you want to change the way a pointer points, you have to assign a different address to it.

• You can find the address you need by looking at other pointers (remember that they store addresses).

Inserting into the Middle of a Linked List

• Suppose we know that there is a Mercedes in a linked list.

• We would like to insert a node containing Honda right after it.

• We first find the Mercedes, using code that we looked at before.

Inserting a Node at Middleelement

maker: Mercedes price: year:operator !=

Node<T> *ptr = start;while ( ptr->info != element ) // element is a parameter

ptr = ptr->next;

After this code executes, ptr points to the node that has Mercedes.

elementmaker: Mercedes price: year:operator !=

Now we would like to insert a CarType object called elementToInsert (containing Honda), which would also be passed in as a parameter, right after the Mercedes

Inserting a Node at Middle (cont.)start ptr

maker: Honda price: 5000year: 1985operator !=

Well, all new nodes are created in the heap, SO…..

elementToInsert

Node<T> *newNode = new Node<T>;

newNode

elementToInsert

Node<T> *newNode = new Node<T>;

newNode

Now, how about placing elementToInsert into the new node?

elementToInsert

Node<T> *newNode = new Node<T>;newNode->info = elementToInsert;

newNode

elementToInsert

newNode

elementToInsert

newNode

Now, what we want is shown by the dashed arrows; this would cause the insertion of the node

newNode

We have two pointers we need to change – but we have to be careful about the way we change them

newNode

If we change the left pointer first, we will no longer be able to access the last node (memory leak)

newNode

So, we first have to assign the address of the last node into the “next” pointer of the new node

newNode

Where is the address of the last node stored?

newNode

That’s right, it is stored in ptr->next

Node<T> *newNode = new Node<T>;newNode->info = elementToInsert;newNode->next = ptr->next;

newNode

Node<T> *newNode = new Node<T>;newNode->info = elementToInsert;newNode->next = ptr->next;ptr->next = newNode;

newNode

Inserting a Node at Middle (cont.)start

Removing a Node

• Suppose we definitely know there is a Mercedes in the linked list and we wish to remove the node that contains it.

• We need to find the node first.

Removing the First Node

Node<T> *ptr = start;if ( ptr->info == element )

start = start->next;

Mercedes is in the first node

Removing the First Node(cont.)

ptr start

Node<T> *ptr = start;if ( ptr->info == element ) {

start = start->next;delete ptr;

startptr

Well, start points to the beginning of the new linked list, but a node isn’t removed unless we free it.

Node<T> *ptr = start;if ( ptr->info == element ) {

start = start->next;delete ptr;

startptr

Now, let’s consider the other case whereby Mercedes is in the middle of the list.

Removing a Middle Node

else {while ( ptr->next->info != element )

ptr = ptr->next;

The while loop points ptr to the node BEFORE the node that has Mercedes.

Removing a Middle Node (cont.)

ptr = ptr->next;

We need to join the node before Mercedes to the node after Mercedes

ptr = ptr->next;

Node<T> *ptr2 = ptr->next;

ptr ptr2

But we mus keep a pointer to the node that has Mercedes

ptr = ptr->next;

ptr->next = ptr2->next;

ptr ptr2

We now join the node before Mercedes to the node after Mercedes

ptr = ptr->next;

Node<T> *ptr2 = ptr->next;ptr->next = ptr2->next;

delete ptr2;

ptr ptr2

We then delete the node that has Mercedes

We did it!

What If Mercedes is the Last Node?

• Would our code still work?• Try to consider every possible situation in

code design.

Removing the Last Node

else {

while ( ptr->next->info != element ) ptr = ptr->next;

Node<T> *ptr2 = ptr->next;ptr->next = ptr2->next;delete ptr2;}

After looping, ptr stops on the next-to-the-last node

ptr = ptr->next;

Node<T> *ptr2 = ptr->next;ptr->next = ptr2->next;delete ptr2;}

start ptr ptr2

Removing the Last Node (cont.)

ptr = ptr->next;

ptr->next = ptr2->next;delete ptr2;}

start ptr ptr2

ptr = ptr->next;

Node<T> *ptr2 = ptr->next;ptr->next = ptr2->next;

delete ptr2;}

start ptr ptr2

The code works in removing the last node.

Working With Linked Lists

• As you can see, sometimes you have to do a lot of thinking and problem-solving when working with linked lists.

• It is not always obvious how to write code.• You can’t memorize the code, because it

will not quite fit situations that you will encounter.

• It is a matter of using logic (and knowing a few tricks of the trade).

Code for Removing a Node

1 Node<T> *ptr = start;2 if ( ptr->info == element ) { 3 start = start->next;4 delete ptr;5 }6 else {7 while ( ptr->next->info != element ) 8 ptr = ptr->next;9 Node<T> *ptr2 = ptr->next;10 ptr->next = ptr2->next;11 delete ptr2;12 }

Here is the resulting code for removing a node. It is assumes the node we are looking for is in the linked list.

• In some situations, an array can be faster than a linked list, such as when a calculated index is used to access an element.

• In other situations, a linked list can be faster than an array, such as when removing an element from the middle (as we saw before).– we usually need to search for the element to remove,

but we search for it in both the array and linked list.

Reference

• Childs, J. S. (2008). Methods for Making Data Structures. C++ Classes and Data Structures. Prentice Hall.

Lecture06 methods for-making_data_structures_v2

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