Skip to main content

Number of nodes in a singly linked list

Now let us write a function to count the number of nodes in a singly linked list.

This one is pretty simple.
  • traverse the list from first node
  • increment count after each node
  • count at the end of traversal is the number of nodes

 int get_count(NODEPTR temp)
 {
      int c = 0;
      while(temp)
      {
    c++;
    temp = temp->next;
      }
     return c; 
}

The function starts with head node and loops till temp becomes NULL. Please remember that NULL has a numerical value of 0 and 0 is false and the condition (temp) is as good as saying (temp!=NULL).

How about a recursive function for the same?


int get_count_recursive(NODEPTR temp)
{
    if(temp!=NULL)
 {
    return get_count_recursive(temp->next)+1;
        }
}

Time to write the driver program for this function

#include<stdio.h>  
#include<stdlib.h>
 struct node  
 {  
   int n;  
   struct node *next;   
 };  
 typedef struct node * NODEPTR;  
  
 NODEPTR create_node(int value)  
 {  
   NODEPTR temp = (NODEPTR) malloc(sizeof(struct node));  
    temp->next = NULL;  
   temp->n = value;  
   return temp;  
 } 

  
 NODEPTR append_node(NODEPTR head, NODEPTR newnode)  
 {  
    NODEPTR temp = head;
    if(temp==NULL)
      return newnode;
    while(temp->next !=NULL)
 temp = temp->next;
    temp->next = newnode;
    return head;  
 } 

 
 void display_nodes(NODEPTR head)  
 {  
   NODEPTR temp = head;//redundant 
   while (temp!= NULL)  
   {  
     printf("%d====>",temp->n);  
     temp = temp->next;  
   }  
   printf("\n");
 } 

 int get_count(NODEPTR temp)
 {
 int c = 0;
      while(temp)
        {
    c++;
    temp = temp->next;
        }
     return c; 
}
int get_count_recursive(NODEPTR temp)
{
    if(temp!=NULL)
 {
    return get_count_recursive(temp->next)+1;
        }
}
 
 int main()  
 {  
     NODEPTR head;  
     NODEPTR newnode,dnode;       
     //initialize head  
     head =  NULL;        
     while(1){
       int value;  
       NODEPTR newnode;  
       printf("node value (-1 to stop)=");  
       scanf("%d",&value); 
       if(value==-1) 
   break; 
       newnode = create_node(value);  
       head = append_node(head,newnode);      
     }
     printf("The linked list now is ");  
     display_nodes(head);
     printf("The number of nodes in sll is %d\n",get_count(head));
     printf("The number of nodes in sll using recursion is %d\n",get_count(head));
           
 }  

Comments

Popular posts from this blog

Introduction to AVL tree

AVL tree is a balanced binary search tree where the difference between heights of two sub trees is maximum 1. Why balanced tree A binary tree is good data structure because search operation here is of the order of O(logn). But this is true if the tree is balanced - which means the left and right subtrees are almost equal in height. If not balanced, search operation will take longer.  In worst case, if the tree has only one branch, then search is of the order O(n). Look at this example.  Here all nodes have only right children.  To search a value in this tree, we need upto 7 iterations, which is O(n). So this tree is very very inefficient. One way of making the tree efficient is to, balance the tree and make sure that height of two branches of each node are almost equal. Height of a node Height of a node is the distance between the node and its extreme child.  In the above a diagram, height of 37 is 3 and height of left child of 37 is 0 and right child of 37 is 2. Bal...

Reverse a singly linked list

One of the commonly used interview question is - how do you reverse a linked list? If you talk about a recursive function to print the list in reverse order, you are so wrong. The question is to reverse the nodes of list. Not print the nodes in reverse order. So how do you go about reversing the nodes. You need to take each node and link it to previous node. But a singly linked list does not have previous pointer. So if n1 is current node, n2 = n1->next, you should set     n2->next = NULL But doing this would cut off the list at n2. So the solution is recursion. That is to reverse n nodes  n1,n2,n3... of a list, reverse the sub list from n2,n3,n4.... link n2->next to n1 set n1->next to NULL The last step is necessary because, once we reverse the list, first node must become last node and should be pointing to NULL. But now the difficulty is regarding the head? Where is head and how do we set it? Once we reach end of list  viz n1->next ==NULL, th...

Balanced brackets

Have you observed something? When ever you are writing code using any IDE, if you write mismatched brackets, immediately an error is shown by IDE. So how does IDE  know if an expression is having balanced brackets? For that, the expression must have equal number of opening and closing brackets of matching types and also they must be in correct order. Let us look at some examples (a+b)*c+d*{e+(f*g)}   - balanced (p+q*[r+u )] - unbalanced (p+q+r+s) ) - unbalanced (m+n*[p+q]+{g+h}) - balanced So we do we write a program to check if an expression is having balanced brackets? We do need to make use of stack to store the brackets. The algorithm is as follows Scan a character - ch from the expression If the character is opening bracket, push it to stack If the character is closing bracket pop a character from stack If popped opening bracket and ch are not of same type ( ( and ) or [ and ] ) stop the function and return false Repeat steps 2 and 3 till all characters are scanned. On...