Chapter 3, Problem 8C

Explanation of Solution

Algorithm:

begin

1.move head of L' list to the head of L llist

2.while(headL ->next !=NULL)

do headL=headL->next

3.headL->next=headM

4. headL' contains the head of both concatinated lists

end

Program code:

//include required header files

#include <iostream>

#include <stdio.h>

#include <stdlib.h>

//use std namespace

using namespace std;

//node structure

struct Node

{

//data memebers

int data;

struct Node *next;

};

//define the method

void linkedlistprint(struct Node *head)

{

//create e node

struct Node *h = head;

//iterate a while loop

while (h != NULL)

{

//print the data

std::cout<<h->data;

//set the h

h = h->next;

}

//new line

std::cout<<"\n";

}

//define a method insert()

void insert(struct Node ** headref, int ndata)

{

//create a new node

struct Node* newnode = (struct Node*) malloc(sizeof(struct Node));

//set the data of the node

newnode->data = ndata;

//set the link of the node

newnode->next = (*headref);

//assign the value

(*headref) = newnode;

}

//define the main() function

int main()

{

//create required variables

struct Node *L = NULL, *M = NULL;

//call the method insert()

insert(&L, 7);

insert(&L, 6);

insert(&L, 5);

//print the statement

printf("L is : ");

//call the method linkedlistprint()

linkedlistprint(L);

//call the method insert()

insert(&M, 4);

insert(&M, 3);

insert(&M, 2...