Skip to main content

Search for a node in Linked List

Normally searching can be done using linear search or binary search. But in a linked list, jumping to an intermediate node is not possible. Hence searching should be done sequentially.

Let us assume we have a linked list of names.

struct node
{
   char name[30];
   struct node * next_ptr;
};
typedef struct node * NODEPTR;


In fact it would be better optimized, if we use a dynamic memory for name. For sake of simplicity, let us leave that. Now we have a node pointer called head, which points to the head of the list. And we also have a string search_string. We need to write a function to search this string, and return the pointer to node containing this string. If search string is not found, we must return NULL. 

NODEPTR head;
-----
-----
NODEPTR search(NODEPTR head, char *str)
{
      NODEPTR temp;
      for(temp=head;temp!=NULL;temp = temp->next_ptr)
      {
             if(strcmp(temp->name,str)==0)
                  return temp;
      }
      return NULL;
}
We start from first node and compare each node till we find the node containing str. But we should not forget to check for end of list, indicated by NULL.

One step where we may go wrong is comparing temp->name with str instead of using strcmp. If we compare them, then the pointers of these two strings are compared, which will never be equal.

Two more ways of optimization are

  1. Not using temp at all. Instead head itself can be used for list traversal. We need not worry that list would get corrupted. It won't as head is a call by value parameter. 
  2. In for condition adding the clause temp->NULL && strcmp(temp->name,str). If strcmp returns 0, condition is false and loop will terminate. But in this case, the body of for loop will be just a semicolon (;). And return statement will be return temp. If we have reached end of list, then temp will be NULL, so we return NULL. 
You can download the complete program from here

Labels

Labels:

Comments

Post a Comment

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...
Post a Comment
Read more

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...
Post a Comment
Read more

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...
2 comments
Read more