unit 3: linked lists part 2: more on linked lists · 1 deletion in singly linked lists (cont’d) 1...

Unit 3: Linked ListsPart 2: More on Linked Lists

Engineering 4892:Data Structures

Faculty of Engineering & Applied ScienceMemorial University of Newfoundland

May 30, 2011

ENGI 4892 (MUN) Unit 3, Part 2 May 30, 2011 1 / 20

1 Deletion in singly linked lists (cont’d)

1 Other Functions

1 Doubly Linked Lists

1 Circular lists

1 Linked lists vs. arrays

Deletion in singly linked lists (cont’d)

We now consider the more general deleteNode method.

deleteNode

deletes the node containing a particular integer (it will delete the firstnode containing the integer, if there are more than one).

First, we will need to search for the node to delete.

Removing a node from an SLL is accomplished by linking the node’spredecessor to its successor.

Like in deleteFromTail we will have to use a loop to find the predecessor.

So we need two loops:

one to find tmp, the node to remove

one to find tmp’s predecessor, pred

Actually, we can combine these into one.

We now consider the more general deleteNode method. deleteNode

In the following, we are trying to delete the 8-node.

The loop begins with pred = head and tmp = head−>next.

At each step, both pred and tmp are incremented:

pred = pred−>next, tmp = tmp−>next

The loop terminates when tmp is at the item to delete: tmp−>info == el.

Actually, if the item is not present this will never happen. So we shouldcheck that tmp != 0 before trying to dereference tmp.

The loop:

IntSLLNode ∗pred , ∗tmp ;for ( pred = head , tmp = head−>next ;

tmp != 0 && ! ( tmp−>info == el ) ;pred = pred−>next , tmp = tmp−>next ) ;

The loop:

Actually, if the item is not present this will never happen.

So we shouldcheck that tmp != 0 before trying to dereference tmp.

The loop:

tmp != 0 && ! ( tmp−>info == el ) ;

pred = pred−>next , tmp = tmp−>next ) ;

The loop:

Before continuing we check that tmp != 0 (if this is not true, the item doesnot exist... so we have nothing to do).

The 8-node is removed by setting pred−>next = tmp−>next.

tmp is delete’ed

This is the code for the general case:

void IntSLList : : deleteNode ( int el ) {. . .IntSLLNode ∗pred , ∗tmp ;for ( pred = head , tmp = head−>next ;

if ( tmp != 0) {pred−>next = tmp−>next ;. . .delete tmp ;

}. . .

There are several special cases:

Trying to delete from an empty list

Could prevent by precondition; Here we choose to allow thisExit immediately if head = 0

}. . .

Could prevent by precondition; Here we choose to allow this

Exit immediately if head = 0

}. . .

Deleting the one node in an SLL of length 1:

Delete the node and set head = tail = 0

if ( head == tail && el == head−>info ) {delete head ;head = tail = 0 ;

Deleting the first node in an SLL of length ≥ 2:

Update head and delete the node

. . . else if (el == head−>info ) {IntSLLNode ∗tmp = head ;head = head−>next ;delete tmp ;

Deleting the last node in an SLL of length ≥ 2:

Same as general case, but set tail = pred

void IntSLList : : deleteNode ( int el ) {if ( head != 0) // i f non−empty l i s t ;

if ( head == tail && el == head−>info ) { // i f on l y one nodedelete head ;head = tail = 0 ;

else if ( el == head−>info ) { // i f more than one nodeIntSLLNode ∗tmp = head ; // and o l d head i s d e l e t e dhead = head−>next ;delete tmp ;

}else { // i f more than one node

IntSLLNode ∗pred , ∗tmp ; // and a non−head d e l e t e dfor ( pred = head , tmp = head−>next ;

if ( tmp != 0) {pred−>next = tmp−>next ;if ( tmp == tail )

tail = pred ;delete tmp ;

}else if ( el == head−>info ) { // i f more than one node

IntSLLNode ∗tmp = head ; // and o l d head i s d e l e t e dhead = head−>next ;delete tmp ;

else { // i f more than one nodeIntSLLNode ∗pred , ∗tmp ; // and a non−head d e l e t e dfor ( pred = head , tmp = head−>next ;

What is the asymptotic complexity of deleteNode?

Best case: O(1)

Worst case: This requires going into our loop n − 1 times. Thus, it isO(n), just like deleteFromTail.

Average case: Assume that every node has an equal chance of beingdeleted. How many iterations through the loop are there for each possibleposition:

First node: 0 iterations (special case)

Second node: 0 iterations

Third node: 1 iteration

Last node: n − 2 iterations

The average number of iterations is:

n(1 + · · ·+ (n − 2)) =

2− 3

n= O(n)

Best case:

n(1 + · · ·+ (n − 2)) =

2− 3

n= O(n)

Best case: O(1)

n(1 + · · ·+ (n − 2)) =

2− 3

n= O(n)

Best case: O(1)

Worst case:

This requires going into our loop n − 1 times. Thus, it isO(n), just like deleteFromTail.

n(1 + · · ·+ (n − 2)) =

2− 3

n= O(n)

Best case: O(1)

n(1 + · · ·+ (n − 2)) =

2− 3

n= O(n)

Best case: O(1)

Average case:

Assume that every node has an equal chance of beingdeleted. How many iterations through the loop are there for each possibleposition:

n(1 + · · ·+ (n − 2)) =

2− 3

n= O(n)

Best case: O(1)

n(1 + · · ·+ (n − 2)) =

2− 3

n= O(n)

Best case: O(1)

n(1 + · · ·+ (n − 2)) =

2− 3

n= O(n)

Best case: O(1)

n(1 + · · ·+ (n − 2)) =

2− 3

n= O(n)

Best case: O(1)

n(1 + · · ·+ (n − 2)) =

2− 3

n= O(n)

Best case: O(1)

n(1 + · · ·+ (n − 2)) =

2− 3

n= O(n)

Best case: O(1)

n(1 + · · ·+ (n − 2)) =

2− 3

n= O(n)

Best case: O(1)

n(1 + · · ·+ (n − 2)) =

2− 3

n= O(n)

Best case: O(1)

n(1 + · · ·+ (n − 2)) =

2− 3

n= O(n)

Best case: O(1)

n(1 + · · ·+ (n − 2)) =

2− 3

Best case: O(1)

n(1 + · · ·+ (n − 2)) =

2− 3

n= O(n)

Other Functions

Consider the following functions of IntSLList,

The search function isInList():

bool IntSLList : : isInList ( int el ) const {IntSLLNode ∗tmp ;for ( tmp = head ; tmp != 0 && ! ( tmp−>info == el ) ;

tmp = tmp−>next ) ;return tmp != 0 ;

Best case: O(1)

Worst case: O(n)

Other Functions

Best case: O(1)

Worst case: O(n)

Other Functions

Best case: O(1)

Worst case: O(n)

Other Functions

Best case: O(1)

Worst case: O(n)

Other Functions

Best case: O(1)

Worst case: O(n)

The destructor:

IntSLList : : ˜ IntSLList ( ) {for ( IntSLLNode ∗p ; ! isEmpty ( ) ; ) {

p = head−>next ;delete head ;head = p ;

The printing function:

void IntSLList : : printAll ( ) const {for ( IntSLLNode ∗tmp = head ; tmp != 0 ; tmp = tmp−>next )

cout << tmp−>info << " " ;cout << endl ;

The destructor:

p = head−>next ;

delete head ;head = p ;

The destructor:

p = head−>next ;delete head ;

head = p ;}

The destructor:

Doubly Linked Lists

Recall that the deleteFromTail function of our SLL was O(n).

We canimprove upon this by introducing doubly linked lists.

Each node in a DLL has two pointers:

One to the previous node in the list: prev

One to the next node in the list: next

At the front of the list the prev pointer is null. Similarly, at the rear next

is null.

Doubly Linked Lists

Recall that the deleteFromTail function of our SLL was O(n). We canimprove upon this by introducing doubly linked lists.

is null.

Doubly Linked Lists

is null.

Doubly Linked Lists

is null.

Doubly Linked Lists

is null.

Doubly Linked Lists

is null.

Doubly Linked Lists

is null.

Doubly Linked Lists

At the front of the list the prev pointer is null.

Similarly, at the rear next

is null.

Doubly Linked Lists

is null.ENGI 4892 (MUN) Unit 3, Part 2 May 30, 2011 13 / 20

The definition of a node is modified to include these pointers, as well as tostore a generic type T.

template<class T>class DLLNode {public :

DLLNode ( ) {next = prev = 0 ;

}DLLNode ( const T& el , DLLNode ∗n = 0 , DLLNode ∗p = 0) {

info = el ; next = n ; prev = p ;}T info ;DLLNode ∗next , ∗prev ;

DLLNode ( const T& el , DLLNode ∗n = 0 , DLLNode ∗p = 0) {info = el ; next = n ; prev = p ;

}T info ;DLLNode ∗next , ∗prev ;

info = el ; next = n ; prev = p ;}

T info ;DLLNode ∗next , ∗prev ;

info = el ; next = n ; prev = p ;}T info ;

DLLNode ∗next , ∗prev ;} ;

info = el ; next = n ; prev = p ;}T info ;DLLNode ∗next , ∗prev ;

Consider the process of inserting into the last position of a doubly-linkedlist.

The following steps are required:

Create a new node:

info is setnext is set to nullprev is set to tail

Modify the current nodes to link in the new one:

tail is set to point at the new nodeThe next member of the former last node is set to point at the newnode

These steps are accomplished by the following member function:

template<class T>void DoublyLinkedList<T> : : addToDLLTail ( const T& el ) {

. . .tail = new DLLNode<T>(el , 0 , tail ) ;tail−>prev−>next = tail ;. . .

Consider the process of inserting into the last position of a doubly-linkedlist. The following steps are required:

Create a new node:

info is setnext is set to

nullprev is set to tail

Create a new node:

info is set

next is set to

nullprev is set to tail

Create a new node:

prev is set to tail

Create a new node:

prev is set to tail

Create a new node:

info is setnext is set to null

prev is set to tail

Create a new node:

info is setnext is set to nullprev is set to

Create a new node:

info is setnext is set to nullprev is set to

Create a new node:

tail is set to point at the new node

The next member of the former last node is set to point at the newnode

Create a new node:

. . .tail = new DLLNode<T>(el , 0 , tail ) ;

tail−>prev−>next = tail ;. . .

Create a new node:

Special case: What if the new node has no predecessor (i.e. the list isempty).

The last step is removed. Also, we have to worry about settinghead.

The complete code for addToDLLTail is as follows:

if ( tail != 0) {tail = new DLLNode<T>(el , 0 , tail ) ;tail−>prev−>next = tail ;

}else head = tail = new DLLNode<T>(el ) ;

Special case: What if the new node has no predecessor (i.e. the list isempty). The last step is removed.

Also, we have to worry about settinghead.

Special case: What if the new node has no predecessor (i.e. the list isempty). The last step is removed. Also, we have to worry about settinghead.

if ( tail != 0) {

tail = new DLLNode<T>(el , 0 , tail ) ;tail−>prev−>next = tail ;

else head = tail = new DLLNode<T>(el ) ;}

Consider deletion of the last node.

This should be much easier than for aSLL because we don’t have to loop to find tail’s predecessor:

For the general case (i.e. list length ≥ 2),

Store the element to delete

Shift tail back by one: tail = tail−>prevDelete last node: delete tail−>nextUpdate the dangling pointer of the last node: tail−>next = 0

Special case: Empty list. Handle this by forbidding it with a precondition.

Special case: One node in list. Delete node and set head = tail = 0

Consider deletion of the last node. This should be much easier than for aSLL because we don’t have to loop to find tail’s predecessor:

Shift tail back by one:

tail = tail−>prevDelete last node: delete tail−>nextUpdate the dangling pointer of the last node: tail−>next = 0

tail = tail−>prev

Delete last node: delete tail−>nextUpdate the dangling pointer of the last node: tail−>next = 0

tail = tail−>prev

Shift tail back by one: tail = tail−>prev

Shift tail back by one: tail = tail−>prevDelete last node:

delete tail−>next

Update the dangling pointer of the last node: tail−>next = 0

Shift tail back by one: tail = tail−>prevDelete last node:

delete tail−>next

Shift tail back by one: tail = tail−>prevDelete last node: delete tail−>next

Shift tail back by one: tail = tail−>prevDelete last node: delete tail−>nextUpdate the dangling pointer of the last node:

tail−>next = 0

Shift tail back by one: tail = tail−>prevDelete last node: delete tail−>nextUpdate the dangling pointer of the last node:

tail−>next = 0

Special case: Empty list.

Handle this by forbidding it with a precondition.

Special case: One node in list.

Delete node and set head = tail = 0

template<class T>T DoublyLinkedList<T> : : deleteFromDLLTail ( ) {

T el = tail−>info ;if ( head == tail ) { // o n l y one node i n l i s t

delete head ;head = tail = 0 ;

else { // more than one nodetail = tail−>prev ;delete tail−>next ;tail−>next = 0 ;

What is the asymptotic complexity of deleteFromDLLTail? O(1)

DLL’s may also tend be more efficient than SLL’s in situations whereoperations occur repeatedly around one part of the list (e.g. text editingwhere each word is represented by a node).

}else { // more than one node

tail = tail−>prev ;delete tail−>next ;tail−>next = 0 ;

What is the asymptotic complexity of deleteFromDLLTail?

Circular lists

In some cases, we may have a list of items that we wish to continuouslycircle through.

e.g. processes sharing some resource such as CPU time.

A circular list (a.k.a circular linked list) would be a suitable data structurefor this purpose,

Circular lists

In some cases, we may have a list of items that we wish to continuouslycircle through. e.g. processes sharing some resource such as CPU time.

Circular lists

Linked lists vs. arrays

We compare arrays with DLL’s.

DLL’s have the best asymptoticperformance for all operations over other LL variants. (Although they maybe less efficient in some cases due to the overhead of additional pointers)

Operation Array Linked list

Indexing 1 O(1) O(n)Search O(n) / O(lg n) 2 O(n)Inserting / Deleting at beginning O(n) O(1)Inserting / Deleting at end O(n) 3 O(1)Inserting / Deleting in middle 4 O(n) O(1)

1Indexing means to access a particular element via an index. Accessing the 100th

element in an array is done by pointer arithmetic. Accessing the 100th node in a linkedlist requires iterating through the first 99 nodes.

2Searching in an ordered array can be done in O(lg n) using binary search.3O(1) if space is available at the end of the array.4Not counting search cost

We compare arrays with DLL’s. DLL’s have the best asymptoticperformance for all operations over other LL variants.

(Although they maybe less efficient in some cases due to the overhead of additional pointers)

We compare arrays with DLL’s. DLL’s have the best asymptoticperformance for all operations over other LL variants. (Although they maybe less efficient in some cases due to the overhead of additional pointers)

Indexing 1 O(1) O(n)

Search O(n) / O(lg n) 2 O(n)Inserting / Deleting at beginning O(n) O(1)Inserting / Deleting at end O(n) 3 O(1)Inserting / Deleting in middle 4 O(n) O(1)

Indexing 1 O(1) O(n)Search O(n) / O(lg n) 2 O(n)

Inserting / Deleting at beginning O(n) O(1)Inserting / Deleting at end O(n) 3 O(1)Inserting / Deleting in middle 4 O(n) O(1)

Indexing 1 O(1) O(n)Search O(n) / O(lg n) 2 O(n)Inserting / Deleting at beginning O(n) O(1)

Inserting / Deleting at end O(n) 3 O(1)Inserting / Deleting in middle 4 O(n) O(1)

Indexing 1 O(1) O(n)Search O(n) / O(lg n) 2 O(n)Inserting / Deleting at beginning O(n) O(1)Inserting / Deleting at end O(n) 3 O(1)

Inserting / Deleting in middle 4 O(n) O(1)

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