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Product of the alternate nodes of linked list


Given with n nodes the task is to print the product of alternate node in a linked list. The program must only print the product of alternate nodes without actually changing the locations of the nodes.

Example

Input -: 10 20 30 40 50 60
Output -: 15000

In the above example, starting from first node which is 10 alternate nodes are 10, 30, 50 and their product is 10*30*50 = 15000.

Product of the alternate nodes of linked list

In the above diagram, blue coloured nodes are the alternate nodes, if we start from first node and red coloured nodes are non considerable nodes.

Approach used below is as follows

  • Take a temporary pointer, lets say, temp of type node

  • Set this temp pointer to first node which is pointed by head pointer

  • Move temp to temp ->next -> next while the situation (temp->next!=NULL && temp!=NULL && temp->next->next!=NULL) holds true

  • Set product=product*(temp->data)

Algorithm

Start
Step 1 -> create structure of a node and temp, next and head as pointer to a structure node
   struct node
      int data
      struct node *next, *head, *temp
   End
Step 2 -> declare function to insert a node in a list
   void insert(int val)
      struct node* newnode = (struct node*)malloc(sizeof(struct node))
      newnode->data = val
      IF head= NULL
         set head = newnode
         set head->next = NULL
      End
      Else
         Set temp=head
         Loop While temp->next!=NULL
            Set temp=temp->next
         End
         Set newnode->next=NULL
         Set temp->next=newnode
      End
Step 3 -> Declare a function to display list
   void display()
      IF head=NULL
         Print no node
   End
Else
   Set temp=head
   Loop While temp!=NULL
      Print temp->data
      Set temp=temp->next
   End
End
Step 4 -> declare a function to find alternate nodes
   void alternate()
      declare int product
      Set temp=head
      Set product=head->data
      Loop While(temp->next!=NULL && temp!=NULL && temp->next-
         >next!=NULL)
         Set temp=temp->next->next
         Set product=product * (temp->data)
      End
      Print product
Step 5 -> in main()
   Create nodes using struct node* head = NULL;
   Call function insert(10) to insert a node
   Call display() to display the list
   Call alternate() to find alternate nodes product
Stop

CODE

#include<stdio.h>
#include<stdlib.h>
//structure of a node
struct node {
   int data;
   struct node *next;
}*head,*temp;
//function for inserting nodes into a list
void insert(int val) {
   struct node* newnode = (struct node*)malloc(sizeof(struct node));
   newnode->data = val;
   if(head == NULL) {
      head = newnode;
      head->next = NULL;
      } else {
      temp=head;
      while(temp->next!=NULL) {
         temp=temp->next;
      }
      newnode->next=NULL;
      temp->next=newnode;
   }
}
//function for displaying a list
void display() {
   if(head==NULL)
      printf("no node ");
   else {
      temp=head;
      while(temp!=NULL) {
         printf("%d ",temp->data);
         temp=temp->next;
      }
   }
}
//function for finding alternate elements
void alternate() {
   int product;
   temp=head;
   product=head->data;
   while(temp->next!=NULL && temp!=NULL && temp->next->next!=NULL) {
      temp=temp->next->next;
      product=product * (temp->data);
   }
   printf("\nproduct of alternate nodes is %d : " ,product);
}
int main() {
   //creating list
   struct node* head = NULL;
   //inserting elements into a list
   insert(10);
   insert(20);
   insert(30);
   insert(40);
   insert(50);
   insert(60);
   //displaying the list
   printf("linked list is : ");
   display();

   //calling alternate function for finding product
   alternate();
   return 0;
}

Output

linked list is : 10 20 30 40 50 60
product of alternate nodes is : 15000