#include #include

typedef struct nod{
int info;
struct nod *next;
}node;

node* SortInsert(node *root, int item); //this function is complete
void simplePrint(node* root); //this function is complete
int countNodes (node* root); //to create this function
node* EvenCopy(node * root); //to create this function

int main()
{
node* head=NULL;
node* head2 = NULL;

node *t;
int ch, ele;
head = SortInsert(head, 4);
head = SortInsert(head,6);
head = SortInsert(head,3);
head = SortInsert(head,5);

printf("\nSimple print List 1: ");
simplePrint(head);

printf("\nCount Nodes %d", countNodes(head)); //modify the countNodes function to make it work

head2 = EvenCopy(head); //modify the EvenCopy function to make it work
printf("\nSimple print after EvenCopy: ");
simplePrint(head2); //This call should print 4, 6

return 0;

}

void simplePrint(node* root)
{
node* t=root;
while(t!=NULL)
{
printf("%d ",t->info);
t=t->next;
}
}

node* SortInsert(node* root, int item)
{
node *temp;
node *t;
temp= (node *) malloc(sizeof(node));
temp->info=item;
temp->next=NULL;
if (root==NULL || root->info >=item)
{
temp->next = root;
root = temp;
}
else
{
t = root;
while (t->next != NULL && t->next->info next;
temp->next = t->next;
t->next = temp;
}

return root;
}

THE FOLLOWING QUESTIONS ARE HERE:

int countNodes (node* root)
{
/*this function takes the head of a linked list and returns the number of nodes available in the linked list. You can use for loops or recursion */

}

node* EvenCopy(node * root)
{
/*this function takes the head of a linked list and copies all the even numbers to a new linked list and return the
head of the new linked list. Note that a number is considered as even if number%2 is 0.
Example: passing the head of a linked list containing 3, 4, 5, 6 would return another linked list containing 4, 6 */

}

I-

Explanation:

#include

#include

typedef struct nod{

int info;

struct nod *next;

}node;

node* SortInsert(node *root, int item); //this function is complete

void simplePrint(node* root); //this function is complete

int countNodes (node* root); //to create this function

node* EvenCopy(node * root); //to create this function

int main()

{

node* head=NULL;

node* head2 = NULL;

node *t;

int ch,ele;

head = SortInsert(head, 4);

head = SortInsert(head,6);

head = SortInsert(head,3);

head = SortInsert(head,5);

printf("\nSimple print List 1: ");

simplePrint(head);

printf("\nCount Nodes %d", countNodes(head)); //modify the countNodes function to make it work

head2 = EvenCopy(head); //modify the EvenCopy function to make it work

printf("\nSimple print after EvenCopy: ");

simplePrint(head2); //This call should print 4, 6

return 0;

}

void simplePrint(node* root)

{

node* t=root;

while(t!=NULL)

{

printf("%d ",t->info);

t=t->next;

}

}

node* SortInsert(node* root, int item)

{

node *temp;

node *t;

temp= (node *) malloc(sizeof(node));

temp->info=item;

temp->next=NULL;

if (root==NULL || root->info >=item)

{

temp->next = root;

root = temp;

}

else

{

t = root;

while (t->next != NULL && t->next->info next;

temp->next = t->next;

t->next = temp;

}

return root;

}

THE FOLLOWING QUESTIONS ARE HERE:

int countNodes (node* root)

{

/*this function takes the head of a linked list and returns the number of nodes available in the linked list. You can use for loops or recursion */

}

node* EvenCopy(node * root)

{

/*this function takes the head of a linked list and copies all the even numbers to a new linked list and return the

head of the new linked list. Note that a number is considered as even if number%2 is 0.

Example: passing the head of a linked list containing 3, 4, 5, 6 would return another linked list containing 4, 6 */

}

you already know many words that you use every day as part of your writing and you probably also know that certain words fit better in certain situations. tonight she will , sit) down and , write) a cover letter to when writing, you need to choose the correct word according to its spelling and

maybe when you are listing people names in a job.

i am not 100% sure