- 76 Views
- Uploaded on
- Presentation posted in: General

Doubly Linked List Lesson xx

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

- Doubly linked list concept
- Node structure
- Insertion sort
- Insertion sort program with a doubly linked list

Head

0

a

b

c

d

e

0

struct node

{

node * prev;

intval;

node * next;

};

7

Ask the use to enter an integer

Store the # in one node of a doubly linked list

Repeat steps 1 & 2 until the user enters ‘s’ to terminate input

Use the insertion sort to sort the #s that are in the doubly linked list

Print out the sorted list

Fig. 1

Head

Head

0

0

2

2

3

5

7

5

8

3

7

8

0

0

Fig. 2

1) Place a pointer called out on the node that we want to insert. (When we 1st start, out is placed on the next to the last node.

2) Place a pointer called in one node to the right of out.

3) Compare the # to be inserted with the contents of what is pointed to by in.

4) If the # to be sorted is less than what is pointed to by in, swap the contents and move in, one node to the right

5) When in is at the end of the list or the # to be sorted is > than what is pointed to in, we have inserted the # into the correct position and it’s time to move out, 1 node to the left and repeat steps 2-5.

We’ll use the insertion sort to rearrange the following list of #s in descending order:

52 7 38

Place a pointer called out on the next to the last #

Place a pointer called in, one node to the right of out

52 7 38

out in

Consider the last # (8) to be the sorted list & 3 is the # we want to insert into the sorted list.

5.If the # pointed to by out (3) < the # pointed to by in, swap them. Now you get the following picture

52 783

out in

After swapping the #s, move in one node to the right

52 783

out in

7. When in is off the list, this means that we have inserted the # 3 in the correct position

8.Move out, 1 node to the left and place in 1 node to the right of out

52 783

out in

9.All the nodes to the right of out are sorted in descending order. Now we are going to insert 7 into the list

10.Since 7 is < 8, we need to swap the numbers and also move in one node to the right.

52 873

out in

11. .Compare 7 and 3. 7 is > 3 so we have inserted 7 in to the correct position in the list.

52 873

out in

12. The #s from out and to the right are now sort in descending order. 8, 7 ,3. Next step is to move out 1 node to the left and place in 1 node to the right of out.

52 873

out in

13. We are going to insert 2 in to the sorted list. You can see that in keeps moving to the right until the # is in the correct position or in is off the list. Out always moves to the left and points to the # we want to insert into the list. This procedure is continues until out points to a null. Then, the list is in descending order.

#include <iostream> using std::cin; using std::cout; using std::flush; using std::endl;

#include <cstdlib>

struct node { node* prev;int value; node* next; };

void printList(const node*);

int main() { char str[15];

node* head = new node; node* tail = head; head‑>prev = 0;

cout<<"enter a number";

cin>>str;

while(str[0]!='s')

{

tail->value=atoi (str);

tail‑>next=new node;

tail‑>next‑>prev=tail;

tail=tail‑>next;

cout<<"enter a number";

cin>>str;

}

tail‑>next=0;

printList(head);//print unsorted list

node* in;

node* out;

int temp;

out=tail‑>prev‑>prev;

while(out!=0)

{

temp=out‑>value;

in=out‑>next;

while(in‑>next!=0&&temp<in‑>value)

{

in‑>prev‑>value=in‑>value;

in‑>value=temp;

in=in‑>next;

}

out=out‑>prev;

}

printList(head); // print list return 0; }

void printList(const node* h) { for (const node* p = h; p; p = p->next) {

cout << "node address: " << p << “ prev “ <<

p->prev << " value " << p->value << " next " << p->next << endl; } }

#include <iostream> using std::cin; using std::cout; using std::flush; using std::endl;

#include <cstdlib>

struct node { node* prev;int value; node* next; };

void printList(const node*);

int main() { char str[15];

node* head = new node; node* tail = head; head‑>prev = 0;

cout<<"enter a number";

cin>>str;

head

0

str

“5”

tail

while(str[0]!='s')

{

tail->value=atoi (str);

tail‑>next=new node;

tail‑>next‑>prev=tail;

tail=tail‑>next;

cout<<"enter a number";

cin>>str;

}

tail‑>next=0;

printList(head);//print unsorted list

head

0

5

str

“5”

tail

while(str[0]!='s')

{

tail->value=atoi (str);

tail‑>next=new node;

tail‑>next‑>prev=tail;

tail=tail‑>next;

cout<<"enter a number";

cin>>str;

}

tail‑>next=0;

printList(head);//print unsorted list

head

0

5

str

“5”

tail

while(str[0]!='s')

{

tail->value=atoi (str);

tail‑>next=new node;

tail‑>next‑>prev=tail;

tail=tail‑>next;

cout<<"enter a number";

cin>>str;

}

tail‑>next=0;

printList(head);//print unsorted list

head

0

5

str

“2”

tail

while(str[0]!='s')

{

tail->value=atoi (str);

tail‑>next=new node;

tail‑>next‑>prev=tail;

tail=tail‑>next;

cout<<"enter a number";

cin>>str;

}

tail‑>next=0;

printList(head);//print unsorted list

0

5

2

7

3

8

0

tail

Head

1) Place a pointer called out on the node that we want to insert. (When we 1st start out is placed on the next to the last node.

2) Place a pointer called in one node to the right of out.

3) Compare the # to be inserted with the contents of what is pointed to by in.

4) If the # to be sorted is less than what is pointed to by in, swap the contents and move in, one node to the right

5) When in is at the end of the list or the # to be sorted is > than what is pointed to in, we have inserted the # into the correct position and it’s time to move out, 1 node to the left and repeat steps 2-5.

while(out!=0)

{

. . .

while(# to be inserted is in the wrong spot)

{

. . .

in=in‑>next; //move in one node to the right

}

out=out‑>prev; //move out one node to the left

}

node* in;

node* out;

int temp;

out=tail‑>prev‑>prev;

Head

0

5

2

7

3

8

0

tail

out

while (out!=0)

{

temp=out‑>value;

in=out‑>next;

while(in‑>next!=0&&temp<in‑>value)

{

in‑>prev‑>value=in‑>value;

in‑>value=temp;

in=in‑>next;

}

out=out‑>prev;

}

temp

3

0

5

2

7

3

8

0

tail

Head

out

in

while (out!=0)

{

temp=out‑>value;

in=out‑>next;

while(in‑>next !=0 && temp < in‑>value)

{

in‑>prev‑>value=in‑>value;

in‑>value=temp;

in=in‑>next;

}

out=out‑>prev;

}

temp

3

0

5

2

7

3

8

0

tail

Head

out

in

while (out!=0)

{

temp=out‑>value;

in=out‑>next;

while(in‑>next !=0 && temp< in‑>value)

{

in‑>prev‑>value=in‑>value;

in‑>value=temp;

in=in‑>next;

}

out=out‑>prev;

}

temp

3

0

5

2

7

8

3

0

tail

Head

in

out

while (out!=0)

{

temp=out‑>value;

in=out‑>next;

while(in‑>next !=0 && temp< in‑>value)

{

in‑>prev‑>value=in‑>value;

in‑>value=temp;

in=in‑>next;

}

out=out‑>prev;

}

temp

7

0

5

2

7

8

3

0

tail

Head

out

in

while (out!=0)

{

temp=out‑>value;

in=out‑>next;

while(in‑>next !=0 && temp< in‑>value)

{

in‑>prev‑>value=in‑>value;

in‑>value=temp;

in=in‑>next;

}

out=out‑>prev;

}

temp

7

0

5

2

8

7

3

0

tail

Head

out

in

while (out!=0)

{

temp=out‑>value;

in=out‑>next;

while(in‑>next !=0 && temp< in‑>value)

{

in‑>prev‑>value=in‑>value;

in‑>value=temp;

in=in‑>next;

}

out=out‑>prev;

}

temp

2

0

5

2

8

7

3

0

tail

Head

out

in

2

temp

Head

1

Head

2

0

0

0

2

2

7

3

5

7

2

8

5

3

5

8

7

3

8

0

0

0

tail

tail

tail

3

out

out

out

in

in

in

5

temp

Head

4

Head

5

0

0

0

7

5

8

7

3

8

2

2

3

2

8

7

5

3

5

0

0

0

tail

tail

tail

Head

6

out

out

in

out

in

in

5

temp

Head

0

8

7

5

3

2

0

tail

out= 0

in

- Doubly linked list concept
- Node structure
- Insertion sort
- Insertion sort program with a doubly linked list