Review on linked lists

Motivation

A “List” is a useful structure to hold a collection of data. Currently, we use arrays for lists

Examples:

List of ten students marksint studentMarks[10];

List of temperatures for the last two weeksdouble temperature[14];

Motivation list using static array

int myArray[1000]; int n;

We have to decide (to oversize) in advance the size of the array (list)

list using dynamic arrayint* myArray; int n;cin >> n;myArray = new int[n];

We allocate an array (list) of any specified size while theprogram is running

linked-list (dynamic size)size = ??The list is dynamic. It can grow and shrink to any size.

Array naturally represents a (ordered) list,

the link is implicit, consecutive and contiguous!

Now the link is explicit, any places!

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Data

Link

20

45

75

85Data Link

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Data Link

0 1 2 array

linked list

Linked Lists: Basic Idea

A linked list is an ordered collection of data Each element of the linked list has

Some data A link to the next element

The link is used to chain the data

Example: A linked list of integers:

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Data Link

The list can grow and shrink

Linked Lists: Basic Ideas

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20 45

addEnd(75), addEnd(85)

deleteEnd(85), deleteHead(20), deleteHead(45)

75

Original linked list of integers:

Insertion (in the middle):

Deletion (in the middle)

Linked Lists: Operations

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60

old value

deleted item

struct Node{ int data;Node* next;

};

We can also:

typedef Node* NodePtr;

Definition of linked list type:

Linked List Structure Node : Data + Link

Definitionstruct Node {

int data; //contains useful information

Node* next; //points to next element or NULL

};

Create a NodeNode* p;

p = new Node; //points to newly allocated memory

Delete a Nodedelete p;

Access fields in a node(*p).data; //access the data field

(*p).next; //access the pointer field

Or it can be accessed this way

p->data //access the data field

p->next //access the pointer field

Representing and accessing linked lists

We define a pointer

Node* head;

that points to the first node of the linked list. When the linked list is empty then head is NULL.

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Passing a Linked List to a Function

When passing a linked list to a function it should suffice to pass the value of head. Using the value of head the function can access the entire list.

Problem: If a function changes the beginning of a list by inserting or deleting a node, then head will no longer point to the beginning of the list.

Solution: When passing head always pass it by reference (not good!)

or using a function to return a new pointer value

It is roughly the same as for an array!!!

Implementation of an (Unsorted) Linked List

Start the first node from scratch

Node* newPtr;

newPtr = new Node;newPtr->data = 20;newPtr->next = NULL; head = newPtr;

Head

newPtr

20

Headhead = NULL;

Inserting a Node at the Beginning

newPtr = new Node;

newPtr->data = 13;

newPtr->next = Head;

head = newPtr;

Head

newPtr

13

20

Keep going …

Head

newPtr

50 40 13 20

void addHead(Node*& head, int newdata){

Node* newPtr = new Node;

newPtr->data = newdata;newPtr->next = Head;head = newPtr;

}

Adding an element to the head:

Call by reference, scaring!!!

NodePtr&

Node* addHead(Node* head, int newdata){

Node* newPtr = new Node;

newPtr->data = newdata;newPtr->next = Head;

return newPtr;}

Also written (more functionally, better!) as:

Compare it with ‘addHead’ with a dynamic array implementation

(to delete)

Deleting the Head Node

Node* p;

p = head;

head = head->next;

delete p;

head

p

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void deleteHead(Node*& head){

if(head != NULL){

NodePtr p = head;

head = head->next;

delete p;

}

}

Node* deleteHead(Node* head){

if(head != NULL){

NodePtr p = head;

head = head->next;

delete p;

}

return head;

}

As a function:

Displaying a Linked List

p = head;

p = p->next;

20 45head

p

20 45head

p

void displayList(Node* head){

NodePtr p;

p = head;

while(p != NULL){

cout << p->data << endl;

p = p->next;

} }

A linked list is displayed by walking through its nodes one by one,

and displaying their data fields (similar to an array!).

void displayArray(int data[], int size) { int n=0; while ( n<size ) {

cout << data[i] << endl; n++;

}

}

For an array:

//return the pointer of the node that has data=item//return NULL if item does not exist

Node* searchNode(Node* head, int item){NodePtr p = head;

NodePtr result = NULL;bool found=false;while((p != NULL) && (!found)){

if(p->data == item) {found = true;result = p;}

p = p->next;}return result;

}

Searching for a node (look at array searching first!)

void main() { const int size=8; int data[size] = { 10, 7, 9, 1, 17, 30, 5, 6 };

int value; cout << "Enter search element: ";

cin >> value; int n=0; int position=-1; bool found=false; while ( (n<size) && (!found) ) {

if(data[n] == value) { found=true; position=n;}

n++;}if(position==-1) cout << "Not found!!\n";else cout << "Found at: " << position << endl;

}

Remember array searching algorithm:

It is essentially the same!

Variations of linked lists

Unsorted linked lists

Sorted linked lists

Circular linked lists Doubly linked lists …

Further considerations for the unsorted lists:

Physical copy of list for operators like ‘delection’ and ‘addHead’

‘delete’ should be understood as a decomposition into a sub-list …

Node* deleteHead(Node* head){

// physically copy head into a new one, newhead

// so to keep the original list intact!

Node* newhead …

if(newhead != NULL){

Node* p = newhead;

newhead = newhead->next;

delete p;

}

return newhead;

}

Original linked list of integers:

Add to the end (insert at the end):

More operation: adding to the end

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Last element

The key is how to locate the last element or node of the list!

void addEnd(NodePtr& head, int newdata){NodePtr newPtr = new Node;newPtr->data = newdata;newPtr->next = NULL;

NodePtr last = head;if(last != NULL){ // general non-empty list case

while(last->next != NULL) last=last->next;

last->next = newPtr;}else // deal with the case of empty list

head = newPtr;}

Add to the end:

Link new object to last->nextLink a new object to empty list

NodePtr addEnd(NodePtr head, int newdata){NodePtr newPtr = new Node;newPtr->data = newdata;newPtr->next = NULL;

NodePtr last = head;if(last != NULL){ // general non-empty list case

while(last->next != NULL) last=last->next;

last->next = newPtr;}else // deal with the case of empty list

head = newPtr;

return head;}

Add to the end as a function:

Implementation of a

Sorted Linked List

Inserting a Node

Head

cur

20

33

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prev

...

newPtr

1. (a) Create a new node using: NodePtr newPtr = new node;

(b) Fill in the data field correctly.

2. Find “prev” and “cur” such that

the new node should be inserted between *prev and *cur.

3. Connect the new node to the list by using:

(a) newPtr->next = cur;

(b) prev->next = newPtr;

Finding prev and cur

Suppose that we want to insert or delete a node with data value newValue. Then the following code successfully finds prev and cur such that

prev->data < newValue <= cur->data

prev = NULL;

cur = head;

found=false;

while( (cur!=NULL) && (!found) ) {

if (newValue > cur->data) {

prev=cur;

cur=cur->next;

}

else found = true;

}

Prev is necessary as we can’t go back!

It’s a kind of search algo,

prev = NULL;

cur = head;

while( (cur!=NULL) && (newValue>cur->data) ) {

prev=cur;

cur=cur->next;

}

Logical AND (&&) is short-circuited, sequential, i.e. if the first part is false, the second part will not be executed.

Finally, it is equivalent to:

//insert item into linked list according to ascending orderNode* insertNode(Node* head, int item){

NodePtr newp, cur, pre; newp = new Node;newp->data = item;

pre = NULL;cur = head;while( (cur != NULL) && (item>cur->data)){

pre = cur;cur = cur->next;

}

if(pre == NULL){ //insert to head of linked listnewp->next = head;head = newp;

} else {pre->next = newp;new->next = cur;

}

return head;}

If the position happens to be the head

General case

// not recommended void type functionvoid insertNode(NodePtr& head, int item){

NodePtr newp, cur, pre; newp = new Node;newp->data = item;

pre = NULL;cur = head;while( (cur != NULL) && (item>cur->data)){

pre = cur;cur = cur->next;

}

if(pre == NULL){ //insert to head of linked listnewp->next = head;head = newp;

} else {pre->next = newp;new->next = cur;

}}

(to delete)

Deleting a Node To delete a node from the list

1. Locate the node to be deleted(a) cur points to the node.

(b) prev points to its predecessor

2. Disconnect node from list using: prev->next = cur->next;

3. Return deleted node to system: delete cur;

Head

cur

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prev

...

Node* deleteNode(Node* head, int item){NodePtr prev=NULL, cur = head;while( (cur!=NULL) && (item > cur->data)){

prev = cur;cur = cur->next;

}

if ( cur!==NULL && cur->data==item) {

if(cur==head)head = head->next;

elseprev->next = cur->next;

delete cur; }

return head;}

Delete an element in a sorted linked list:

If the element is at the head

General case

We can delete only if the element is present!

If (cur==NULL || cur->data!=item) Item is not in the list!

Get the location

void deleteNode(NodePtr& head, int item){NodePtr prev=NULL, cur = head;while( (cur!=NULL) && (item > cur->data)){

prev = cur;cur = cur->next;

}

if ( cur!==NULL && cur->data==item) {

if(cur==Head)Head = Head->next;

elseprev->next = cur->next;

delete cur; }}

// in a void function, not recommended

If the element is at the head

General case

We can delete only if the element is present!

If (cur==NULL || cur->data!=item) Item is not in the list!

Get the location