Use C++ Write a function to merge two doubly linked lists. The input lists have their elements in
sorted order, from lowest to highest. The output list should also be sorted from lowest to highest.
Your algorithm should run in linear time on the length of the output list. Provide an algorithm for
your function Implement and show some samples of your running function
Solution
/* This is a C++ program to merge two sorted linked lists and produce a list in a
Sorted order */
#include
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* This function is used to pull off the front node of the source and put it in destnation */
void letusMoveNode(struct node** destRef, struct node** sourceRef);
/* This is a function used for performing merging of two linked lists.It takes two lists sorted in
increasing order, and splices
their nodes together to make one big sorted list which
is returned. */
struct node* letssortlists(struct node* a, struct node* b)
{
/* Let us first have dummyone first node to hang the result on */
struct node dummyone;
/*next the tail points to the last result node */
struct node* tail = &dummyone;
/* As a result, tail->next is the place to add new nodes
to the result. */
dummyone.next = NULL;
while (1)
{
if (a == NULL)
{
/*we will check that if either list runs out, use the
other list */
tail->next = b;
break;
}
else if (b == NULL)
{
tail->next = a;
break;
}
if (a->data <= b->data)
letusMoveNode(&(tail->next), &a);
else
letusMoveNode(&(tail->next), &b);
tail = tail->next;
}
return(dummyone.next);
}
/* letusMoveNode() function is a function which takes the node from the front of the
source, and move it to the front of the dest.
Before calling letusMoveNode():
source == {1, 2, 3}
dest == {1, 2, 3}
Affter calling letusMoveNode():
source == {3, 4}
dest == {1, 2, 3, 4} */
void letusMoveNode(struct node** destReference, struct node** sourceReference)
{
/*This is the front source node */
struct node* newNode = *sourceReference;
assert(newNode != NULL);
/* now we will advance the source pointer */
*sourceReference = newNode->next;
/*next we will link the old dest off the new node */
newNode->next = *destReference;
/* Fineally we will move dest to point to the new node */
*destReference = newNode;
}
/* This IS A function which is used to insert a node at the beginning of the linked list */
void insertpush(struct node** head_reference, int new_data)
{
/* This is used to allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* now we will put in the data */
new_node->data = new_data;
/* The next step is to link the old list off the new node */
new_node->next = (*head_reference);
/*Finally we will move the head to point to the new node */
(*head_reference) = new_node;
}
/* This is a function to print nodes in a given linked list */
void printList(struct node *node)
{
while (node!=NULL)
{
printf(\"%d \", node->data);
node = node->next;
}
}
/*This is the driver program to test ab.
Use C++ Write a function to merge two doubly linked lists. The input.pdf
1. Use C++ Write a function to merge two doubly linked lists. The input lists have their elements in
sorted order, from lowest to highest. The output list should also be sorted from lowest to highest.
Your algorithm should run in linear time on the length of the output list. Provide an algorithm for
your function Implement and show some samples of your running function
Solution
/* This is a C++ program to merge two sorted linked lists and produce a list in a
Sorted order */
#include
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* This function is used to pull off the front node of the source and put it in destnation */
void letusMoveNode(struct node** destRef, struct node** sourceRef);
/* This is a function used for performing merging of two linked lists.It takes two lists sorted in
increasing order, and splices
their nodes together to make one big sorted list which
is returned. */
struct node* letssortlists(struct node* a, struct node* b)
{
/* Let us first have dummyone first node to hang the result on */
struct node dummyone;
/*next the tail points to the last result node */
struct node* tail = &dummyone;
/* As a result, tail->next is the place to add new nodes
to the result. */
dummyone.next = NULL;
while (1)
{
2. if (a == NULL)
{
/*we will check that if either list runs out, use the
other list */
tail->next = b;
break;
}
else if (b == NULL)
{
tail->next = a;
break;
}
if (a->data <= b->data)
letusMoveNode(&(tail->next), &a);
else
letusMoveNode(&(tail->next), &b);
tail = tail->next;
}
return(dummyone.next);
}
/* letusMoveNode() function is a function which takes the node from the front of the
source, and move it to the front of the dest.
Before calling letusMoveNode():
source == {1, 2, 3}
dest == {1, 2, 3}
Affter calling letusMoveNode():
source == {3, 4}
dest == {1, 2, 3, 4} */
void letusMoveNode(struct node** destReference, struct node** sourceReference)
{
/*This is the front source node */
struct node* newNode = *sourceReference;
assert(newNode != NULL);
/* now we will advance the source pointer */
*sourceReference = newNode->next;
/*next we will link the old dest off the new node */
3. newNode->next = *destReference;
/* Fineally we will move dest to point to the new node */
*destReference = newNode;
}
/* This IS A function which is used to insert a node at the beginning of the linked list */
void insertpush(struct node** head_reference, int new_data)
{
/* This is used to allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* now we will put in the data */
new_node->data = new_data;
/* The next step is to link the old list off the new node */
new_node->next = (*head_reference);
/*Finally we will move the head to point to the new node */
(*head_reference) = new_node;
}
/* This is a function to print nodes in a given linked list */
void printList(struct node *node)
{
while (node!=NULL)
{
printf("%d ", node->data);
node = node->next;
}
}
/*This is the driver program to test above functions*/
int main()
{
/* let us start with the empty list */
struct node* result = NULL;
struct node* a = NULL;
struct node* b = NULL;
/* Firstly we will create two sorted linked lists to test
the functions
Created lists, a: 5->12->15, b: 2->4->18 */
4. insertpush(&a, 15);
insertpush(&a, 12);
insertpush(&a, 5);
push(&b, 20);
push(&b, 3);
push(&b, 2);
/* now we will be removing duplicates from linked list */
res = letssortlists(a, b);
printf("Merged Linked List is: ");
printList(result);
return 0;
}
OUTPUT:
Merged linked list is:
2 3 5 12 15 18
