The document describes creating two linked lists with eight nodes each. The lists are sorted as they are created with each node containing a last name and age. The lists are then printed. Nodes under age 40 are deleted from the first list and nodes over age 50 are deleted from the second list. The remaining nodes from both lists are then merged into a single sorted list.

1. Create two linked lists each having eight nodes…..the first list has a pointer called UNO to the
first node and the second list has a pointer called DUO to the first node. The lists are to be sorted
as you create them. The data in the lists will consist of a last name and age (e.g. Pinto 39)..now
don’t laugh!!! Print out each sorted list after you create them…..then delete all nodes from the
FIRST list whose age is under 40…and delete all nodes from the second list whose age is over
50. Print out both Finally..merge the REMAINING nodes from both lists into one list..headed by
first. Print out the final list.
Solution
#include
#include
#include
using namespace std;
void getdata(string& lastname,Int&age);
const int nil = 0;
class node_type
{
public:
string lname;
Int lage;
node_type *next;
};
void main()
{
node_type *uno, *duo, *p, *r, *newnode, *newnode2;
int i;
int age_value;
string lastname;
uno = new node_type;
p = uno;
getdata(lastname,age_value);
(*uno).lname = lastname;
(*uno).lage = age_value;
(*uno).next = nil;

2. duo = new node_type;
r = duo;
(*duo).lname = lastname;
(*duo).lage = age_value;
(*duo).next = nil;
for (i=0; i<=8;++i)
{
getdata(lastname,age_value);
newnode = new node_type;
newnode2 = new node_type;
(*newnode).lname = lastname;
(*newnode).lage = age_value;
(*newnode).next = nil;
(*newnode2).name = lastname;
(*newnode2).lage = age_value;
(*newnode2).next = nil;
(*p).next = newnode;
(*r).next = newnode2;
p = newnode;
r = newnode2;
}
}
void getdata(string& lastname,float& age_value)
{
cout << "Enter last name " ;
cin >> lastname;
cout << lastname << " ";
cout << "Enter age " ;
cin >> age_value;
cout << age_value << " ";
}
int sort[5];
int temp,i,j;
clrscr();
printf("Enter list of array: ");
for(i=0;i<5;i++){

3. scanf("%d",&sort[i]);
}
Now for sorting of the linked lists
/* Sorting for loop */
for(i=0;i<3;i++){
for(j=0;j<3;j++){
if(sort[i] > sort[j]){
temp=sort[j];
sort[j]=sort[i];
sort[i]=temp;
}
}
/* end of sorting */
for(i=0;i<5;i++){
printf("%d",&sort[i]);
}
For merging the linked
{
for(temp1 = uno; temp1->next!=NULL; temp1 = temp1->next);
if ((*temp1).next = NULL)
{
(*temp1).next = duo;
}
}
That should leave temp1 pointing to the last node. Then assign
temp1->next = duo;