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# Doubly Linked List Lesson xx - PowerPoint PPT Presentation

Doubly Linked List Lesson xx. Objectives. Doubly linked list concept Node structure Insertion sort Insertion sort program with a doubly linked list. Illustration of a Doubly Linked List. Head. 0. a. b. c. d. e. 0. Node Structure for a Doubly Linked List Node. struct node {

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Presentation Transcript

• Node structure

• Insertion sort

• Insertion sort program with a doubly linked list

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

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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:

5 2 7 3 8

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

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

5 2 7 3 8

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

5 2 7 8 3

out in

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

5 2 7 8 3

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

5 2 7 8 3

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.

5 2 8 7 3

out in

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

5 2 8 7 3

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.

5 2 8 7 3

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];

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;

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];

cout<<"enter a number";

cin>>str;

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;

0

5

str

“5”

tail

Connect 2 Prototypend Node to 1st

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;

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;

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;

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2

7

3

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0

tail

Insertion Sort Logic Prototype

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.

Basic Code Outline Prototype

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

}

Set Up Pointers for Insertion Sort Prototype

node* in;

node* out;

int temp;

out=tail‑>prev‑>prev;

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tail

out

Set Up in & temp Prototype

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

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tail

out

in

See If # is in Correct Position Prototype

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

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tail

out

in

Swap Prototype

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

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tail

in

out

Insert Next # 7 into List Prototype

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

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tail

out

in

Inner Loop the 2 Prototypend Time

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

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tail

out

in

Insert Next # 2 into List Prototype

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

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tail

out

in

2 Prototype

temp

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tail

tail

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out

out

out

in

in

in

5 Prototype

temp

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out

out

in

out

in

in

5 Prototype

temp

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out= 0

in

Summary Prototype