chapter 6 lists plus

Chapter 6

Lists Plus

Lecture 12

What is a Circular Linked List?

• A circular linked list is a list in which every node has a successor; the “last” element is succeeded by the “first” element.

External Pointer to the Last Node

What is a Doubly Linked List?

• A doubly linked list is a list in which each node is linked to both its successor and its predecessor.

Linking the New Node into the List

Deleting from a Doubly Linked List

What are the advantages of a circular doubly linked list?

What are Header and Trailer Nodes?

• A Header Node is a node at the beginning of a list that contains a key value smaller than any possible key.

• A Trailer Node is a node at the end of a list that contains a key larger than any possible key.

• Both header and trailer are placeholding nodes used to simplify list processing.

A linked list in static storage ?

A Linked List as an Array of Records

A Sorted list Stored in an Array of Nodes

An Array with Linked List of Values and Free Space

An Array with Three Lists (Including the Free List)

What is a Class Template?

• A class template allows the compiler to generate multiple versions of a class type by using type parameters.

• The formal parameter appears in the class template definition, and

• the actual parameter appears in the client code.

• Both are enclosed in pointed brackets, < >

//--------------------------------------------------------// CLASS TEMPLATE DEFINITION//--------------------------------------------------------#include "ItemType.h" // for MAX_ITEMS and ItemType

template<class ItemType> // formal parameter listclass StackType {public:

StackType( ); bool IsEmpty( ) const;bool IsFull( ) const;void Push( ItemType item );void Pop( ItemType item );ItemType Top( );

private:int top;ItemType items[MAX_ITEMS];

};

StackType<int> numStack;

top 3

[MAX_ITEMS-1] . . .

[ 3 ] 789

[ 2 ] -56

[ 1 ] 132

items [ 0 ] 5670

ACTUAL PARAMETER

top 3

[MAX_ITEMS-1] . . .

[ 3 ] 3456.8

[ 2 ] -90.98

[ 1 ] 98.6

items [ 0 ] 167.87

StackType<float> numStack;ACTUAL PARAMETER

top 3

[MAX_ITEMS-1] . . .

[ 3 ] Bradley

[ 2 ] Asad

[ 1 ] Rodrigo

items [ 0 ] Max

StackType<string> numStack;ACTUAL PARAMETER

//--------------------------------------------------------// SAMPLE CLASS MEMBER FUNCTIONS //--------------------------------------------------------

template<class ItemType> // formal parameter listStackType<ItemType>::StackType( ){

top = -1;}

template<class ItemType> // formal parameter list void StackType<ItemType>::Push ( ItemType newItem ){

if (IsFull()) throw FullStack(); top++;

items[top] = newItem; // STATIC ARRAY IMPLEMENTATION

}

Using class templates

The actual parameter to the template is a data type. Any type can be used, either built-in or user-defined.

Templates are instantiated at run time.

Can you see how this might cause a problem?

When creating class template

• Put .h and .cpp code in the same .h file

or

• Have .h include .cpp file at the end

Recall Definition of Stack

• Logical (or ADT) level: A stack is an ordered group of homogeneous items (elements),

• in which the removal and addition of stack items can take place only at the top of the stack.

• A stack is a LIFO “last in, first out” structure.

class StackType<int>

StackType

Top

Pop

Push

IsFull

IsEmpty Private data:

topPtr

~StackType

20 30

What happens . . .

• When a function is called that uses pass by value for a class object like our dynamically linked stack?

23

Pass by value makes a shallow copy

20 30

StackType<int> myStack; // CLIENT CODE . . .

MyFunction( myStack ); // function call

Private data: 7000 6000

topPtr 7000

Private data:

topPtr 7000

myStack

shallow copy

momeStack

What’s the difference?

• A shallow copy shares the pointed to data with the original class object.

• A deep copy stores its own copy of the pointed to data at different locations than the data in the original class object.

Making a deep copy

20 30

Private data: 7000 6000

topPtr 7000someStack

20 30

Private data: 5000 2000

topPtr 5000

myStack

deep copy

// FUNCTION CODE

template<class ItemType>

void MyFunction( StackType<ItemType> SomeStack )

// Uses pass by value

{

ItemType item;

SomeStack.Pop(item); .

.

.

}

Suppose MyFunction Uses Pop

What happens in a shallow copy?

27

MyStack.topPtr is left dangling

? 30

StackType<int> myStack; // CLIENT CODE . . .

MyFunction( myStack );

Private data:

topPtr 6000

myStack someStack

shallow copy

Private data: 7000 6000

topPtr 7000

28

MyStack.topPtr is left dangling

? 30

Private data:

topPtr 6000

myStack someStack

shallow copy

Private data: 7000 6000

topPtr 7000

Notice that the shallow copy and the actual parameter myStack, have changed!

As a result . . .

• This default method used for pass by value is not the best way when a data member points to dynamic data.

• Instead, you should write what is called a copy constructor, • which makes a deep copy of the dynamic data in

a different memory location.

More about copy constructors

• When there is a copy constructor provided for a class, the copy constructor is used to make copies for pass by value.

• You do not call the copy constructor.

• Like other constructors, it has no return type.

• Because the copy constructor properly defines pass by value for your class, it must use pass by reference in its definition.

// DYNAMICALLY LINKED IMPLEMENTATION OF STACK template<class ItemType>class StackType {public: StackType( );

// Default constructor.// Post: Stack is created and empty.

StackType( const StackType<ItemType>& anotherStack );// Copy constructor.// Implicitly called for pass by value.. ..

~StackType( ); // Destructor.// Post: Memory for nodes has been deallocated.

private:NodeType<ItemType>* topPtr;

};

CLASS CONSTRUCTOR

CLASS COPY CONSTRUCTOR

CLASS DESTRUCTOR

Classes with Data Member Pointers Need

template<class ItemType> // COPY CONSTRUCTORStackType<ItemType>::StackType( const StackType<ItemType>& anotherStack ){ NodeType<ItemType>* ptr1;

NodeType<ItemType>* ptr2;if ( anotherStack.topPtr == NULL )

topPtr = NULL;else // allocate memory for first node{ topPtr = new NodeType<ItemType>;

topPtr->info = anotherStack.topPtr->info;ptr1 = anotherStack.topPtr->next;ptr2 = topPtr;while ( ptr1 != NULL ) // deep copy other nodes{ ptr2->next = new NodeType<ItemType>;

ptr2 = ptr2->next;ptr2->info = ptr1->info;ptr1 = ptr1->next;

}ptr2->next = NULL;

}}

What About the Assignment Operator?

• The default method used for assignment of class objects makes a shallow copy.

• If your class has a data member pointer to dynamic data, • you should write a member function to overload

the assignment operator to make a deep copy of the dynamic data.

// DYNAMICALLY LINKED IMPLEMENTATION OF STACK template<class ItemType>class StackType {public:

StackType( ); // Default constructor.StackType(const StackType<ItemType>& anotherStack );// Copy constructor.void operator= ( StackType<ItemType> );// Overloads assignment operator.. ..

~StackType( ); // Destructor.

private:NodeType<ItemType>* topPtr;

};

Using Overloaded Binary operator+

When a Member Function was definedmyStack + yourStack

myStack.operator+(yourStack)

When a Friend Function was defined

myStack + yourStack

operator+(myStack, yourStack)

C++ Operator Overloading Guides

1 All operators except these :: . sizeof ?: may be overloaded.

2 At least one operand must be a class instance.

3 You cannot change precedence, operator symbols, or number of operands.

4 Overloading ++ and -- requires prefix form use by default.

6 An operator can be given multiple meanings if the data types of operands differ.

Polymorphism with Virtual Functions

formalParameter.MemberFunction(...);then1.If MemberFunction is not a virtual function, the type of the formal parameter determines which function to call.

• Static binding is used.

2.If MemberFunction is a virtual function, the type of the actual parameter determines which function to call.

• Dynamic binding is used.

class ItemType { public: virtual RelationType ComparedTo(ItemType) const; private: char lastName[50] ;} ;class NewItemType : public ItemType { public: RelationType ComparedTo( ItemType) const; private: char firstName[50] ;} ;void PrintResult( ItemType& first, ItemType& second) { using namespace std; if ( first. ComparedTo( second) ==LESS) cout << " First comes before second" ; else cout << " First does not come before second" ;}

ItemType item1 , item2;NewItemType item3 , item4:PrintResult( item1 , item2) ;PrintResult( item3 , item4) ;

You should pass by reference to access derived data members

Chapter 7Programming with

Recursion

What Is Recursion?

• Recursive call A method call in which the method being called is the same as the one making the call

• Direct recursion Recursion in which a method directly calls itself

• Indirect recursion Recursion in which a chain of two or more method calls returns to the method that originated the chain

Recursion

• You must be careful when using recursion.

• Recursive solutions are typically less efficient than iterative solutions.

• Still, many problems lend themselves to simple, elegant, recursive solutions.

• We must avoid making an infinite sequence of function calls

• infinite recursion

Some Definitions

• Base case The case for which the solution can be stated nonrecursively

• General (recursive) case The case for which the solution is expressed in terms of a smaller version of itself

• Recursive algorithm A solution that is expressed in terms of

(a)smaller instances of itself and (b)a base case

Writing a recursive function to find n factorial

DISCUSSION

The function call Factorial(4) should have value 24, because that is 4 * 3 * 2 * 1 .

For a situation in which the answer is known, the value of 0! is 1.

So our base case could be along the lines of

if ( number == 0 )

return 1;

General format for many recursive functions if (some condition for which answer is known)

// base case

solution statement

else // general case

recursive function call

• Each successive recursive call should bring you closer to a situation in which the answer is known.

• Each recursive algorithm must have at least one base case, as well as the general (recursive) case

Writing a recursive function to find Factorial(n)

Now for the general case . . .

The value of Factorial(n) can be written as

n * the product of the numbers from (n - 1) to 1,

that is,

n * (n - 1) * . . . * 1

or,

n * Factorial(n - 1)

And notice that the recursive call Factorial(n - 1) gets us “closer” to the base case of Factorial(0).

Recursive Solution

int Factorial ( int number )

// Pre: number is assigned and number >= 0.

{

if ( number == 0) // base case

return 1 ;

else // general case

return number + Factorial ( number - 1 ) ;

}

Three-Question Method of verifying recursive functions

• Base-Case Question: Is there a nonrecursive way out of the function?

• Smaller-Caller Question: Does each recursive function call involve a smaller case of the original problem leading to the base case?

• General-Case Question: Assuming each recursive call works correctly, does the whole function work correctly?

Recursive function to determine if value is in list

PROTOTYPE

bool ValueInList( ListType list , int value , int startIndex ) ;

Already searched Needs to be searched

74 36 . . . 95

list[0] [1] [startIndex]

75 29 47 . . .

[length -1]

index of currentelement to examine

bool ValueInList ( ListType list , int value, int startIndex )

// Searches list for value between positions startIndex// and list.length-1// Pre: list.info[ startIndex ] . . list.info[ list.length - 1 ]// contain values to be searched// Post: Function value = // ( value exists in list.info[ startIndex ] . . // list.info[ list.length - 1 ] ){ if ( list.info[startIndex] == value ) // one base case

return true ; else if (startIndex == list.length -1 ) // another base case

return false ;else // general casereturn ValueInList( list, value, startIndex + 1 ) ;

}

“Why use recursion?”

Those examples could have been written without recursion, using iteration instead.

The iterative solution uses a loop, and the recursive solution uses an if statement.

However, for certain problems the recursive solution is the most natural solution.

This often occurs with pointer variables.

An example where recursion comes naturally

• Combinations• how many combinations of a certain

size can be made out of a total group of elements

int Combinations(int group, int members)// Pre: group and members are positive. group >= members// Post: Function value = number of combinations of members // size that can be constructed from the total group size.{ if (members == 1) return group; // Base case 1

else

if (members == group) return 1; // Base case 2

else return (Combinations(group-1, members-1) +

Combinations(group-1, members) ) ;}

Combinations(4, 3)

struct NodeType{

int info ;NodeType* next ;

}

class SortedType {public :

. . .// member function prototypes

private :

NodeType* listData ;

} ;

struct ListType

RevPrint(listData);

A B C D E

FIRST, print out this section of list, backwards

THEN, print this element

listData

61

Using recursion with a linked list

void RevPrint ( NodeType* listPtr )

// Pre: listPtr points to an element of a list.

// Post: all elements of list pointed to by listPtr

// have been printed out in reverse order.

{

if ( listPtr != NULL ) // general case

{

RevPrint ( listPtr-> next ); // process the rest

std::cout << listPtr->info << std::endl;

// print this element

}

// Base case : if the list is empty, do nothing

}

Function BinarySearch( )

• BinarySearch takes sorted array info, and two subscripts, fromLoc and toLoc, and item as arguments. • It returns false if item is not found in the

elements info[fromLoc…toLoc]. • Otherwise, it returns true.

• BinarySearch can be written using iteration, or using recursion.

found = BinarySearch(info, 25, 0, 14 );

item fromLoc toLocindexes

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

info 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28

16 18 20 22 24 26 28

24 26 28

24 NOTE: denotes element examined

template<class ItemType>bool BinarySearch ( ItemType info[ ] , ItemType item , int fromLoc , int toLoc )

// Pre: info [ fromLoc . . toLoc ] sorted in ascending order // Post: Function value = ( item in info [ fromLoc .. toLoc] )

{ int mid ; if ( fromLoc > toLoc ) // base case -- not found return false ; else {

mid = ( fromLoc + toLoc ) / 2 ; if ( info [ mid ] == item ) //base case-- found at mi return true ;

else if ( item < info [ mid ] ) // search lower half return BinarySearch ( info, item, fromLoc, mid-1 ) ; else // search upper half

return BinarySearch( info, item, mid + 1, toLoc ) ; } }

Static allocation of space

Recursion is not possible

variable names should be bound to actual addressesin memory at run time

When a function is called...

• A transfer of control occurs from the calling block to the code of the function.

• It is necessary that there be a return to the correct place in the calling block after the function code is executed.

• This correct place is called the return address.

• When any function is called, the run-time stack is used. • On this stack is placed an activation record (stack frame)

for the function call.

Stack Activation Frames

• The activation record stores • the return address for this function call, • the parameters, • local variables, and • the function’s return value, if non-void.

• The activation record for a particular function call is popped off the run-time stack when

• the final closing brace in the function code is reached, or • when a return statement is reached in the function code.

• At this time the function’s return value, if non-void, is brought back to the calling block return address for use there.

// Another recursive function int Func ( int a, int b )

// Pre: a and b have been assigned values// Post: Function value = ??

{ int result;

if ( b == 0 ) // base caseresult = 0;

else if ( b > 0 ) // first general caseresult = a + Func ( a , b - 1 ) ); // instruction 50

else // second general caseresult = Func ( - a , - b ); // instruction 70

return result;}

What operation does Func(a, b) simulate?

FCTVAL ? result ? b 2 a 5Return Address 100

x = Func(5, 2); // original call is instruction 100

original call at instruction 100 pushes on this record for Func(5,2)

Run-Time Stack Activation Records

FCTVAL ? result ? b 1 a 5Return Address 50

FCTVAL ? result 5+Func(5,1) = ? b 2 a 5Return Address 100

record for Func(5,2)

call in Func(5,2) codeat instruction 50 pushes on this recordfor Func(5,1)

x = Func(5, 2); // original call at instruction 100

Run-Time Stack Activation Records

FCTVAL ? result ? b 0 a 5Return Address 50

FCTVAL ? result 5+Func(5,0) = ? b 1 a 5Return Address 50

FCTVAL ? result 5+Func(5,1) = ? b 2 a 5Return Address 100

record for Func(5,2)

record for Func(5,1)

call in Func(5,1) codeat instruction 50pushes on this record for Func(5,0)

x = Func(5, 2); // original call at instruction 100 Run-Time Stack Activation Records

FCTVAL 0 result 0 b 0 a 5Return Address 50

FCTVAL ? result 5+Func(5,0) = ? b 1 a 5Return Address 50

FCTVAL ? result 5+Func(5,1) = ? b 2 a 5Return Address 100

record for Func(5,2)

record for Func(5,1)

record for Func(5,0)is popped first with its FCTVAL

x = Func(5, 2); // original call at instruction 100

Run-Time Stack Activation Records

FCTVAL 5 result 5+Func(5,0) = 5+ 0 b 1 a 5Return Address 50

FCTVAL ? result 5+Func(5,1) = ? b 2 a 5Return Address 100

record for Func(5,2)

record for Func(5,1)is popped nextwith its FCTVAL

x = Func(5, 2); // original call at instruction 100

Run-Time Stack Activation Records

FCTVAL 10 result 5+Func(5,1) = 5+5 b 2 a 5Return Address 100

x = Func(5, 2); // original call at line 100

record for Func(5,2)is popped lastwith its FCTVAL

Run-Time Stack Activation Records

Tail Recursion

• The case in which a function contains only a single recursive call and it is the last statement to be executed in the function.

• Tail recursion can be replaced by iteration to remove recursion from the solution as in the next example.

// USES TAIL RECURSION

bool ValueInList ( ListType list , int value , int startIndex )

// Searches list for value between positions startIndex

// and list.length-1

// Pre: list.info[ startIndex ] . . list.info[ list.length - 1 ]

// contain values to be searched

// Post: Function value =

// ( value exists in list.info[ startIndex ] . .

// list.info[ list.length - 1 ] )

{

if ( list.info[startIndex] == value ) // one base case return true;

else if (startIndex == list.length -1 ) // another base case

return false;

else

return ValueInList( list, value, startIndex + 1 );

}

// ITERATIVE SOLUTION

bool ValueInList ( ListType list , int value , int startIndex )

// Searches list for value between positions startIndex// and list.length-1// Pre: list.info[ startIndex ] . . list.info[ list.length - 1 ]// contain values to be searched// Post: Function value = // ( value exists in list.info[ startIndex ] . . // list.info[ list.length - 1 ] ){ bool found = false; while ( !found && startIndex < list.length )

{ if ( value == list.info[ startIndex ] ) found = true;

else startIndex++;}return found;

}

Use a recursive solution when:

• The depth of recursive calls is relatively “shallow” compared to the size of the problem.

• The recursive version does about the same amount of work as the nonrecursive version.

• The recursive version is [much] shorter and simpler than the nonrecursive solution.

SHALLOW DEPTH EFFICIENCY CLARITY