The Problem Complete the function printList that accepts a node * (the head of a

Question

The Problem

Complete the function printList that accepts a node * (the head of a linked list) as a parameter and print out the list in the following format:

If the list is empty then the function should print "Empty list".

You should complete the main function to test your printList in, at least, the following cases:

An empty list

A list with exactly one node

A list with more than one node (eg: 100 -> 200 -> 300, as shown above)

//main.cpp

#include <iostream>

using namespace std;

struct node {

int val;

node *next;

};

void printList(node *head) {

// printList function

}

int main() {

// Test 1: An empty list

node * head = NULL;

printList(head);

// Test 2: A list with exactly one node

// Test 3: A list with more than one node

return 0;

}

Explanation / Answer

#include <iostream>
using namespace std;

struct node {
   int val;
    node *next;
};

void printList(node *head)
{
int i = 0;
if(head == NULL)   // if head is NUll it is empty list
{
    cout<<" Empty list";
}
while(head != NULL) //while end of the list is not reached
{
   cout<<" Node "<<i<<":"<<head->val;
   head = head->next;   // traverse next node
}

}

int main() {

// Test 1: An empty list

cout<<" list with no node";
node *head = NULL;

printList(head);

// Test 2: A list with exactly one node

cout<<" list with one node";
node *node1 = new node;    //make new node
node1->val= 100;   //assign its value to 100
node1->next = NULL; //assign its next pointer to null
head= node1;   // make it the head node as head is null
printList(head); // call printList function

// Test 3: A list with more than one node
cout<<" list with three nodes";
node *node2 = new node;
node2->val= 200;
node2->next = NULL;
node1->next = node2;


node *node3 = new node;
node3->val= 300;
node3->next = NULL;
node2->next = node3;
printList(head);

return 0;

}

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