Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

C Program For Searching An Element In A Linked List

  • Last Updated : 27 Dec, 2021

Write a function that searches a given key ‘x’ in a given singly linked list. The function should return true if x is present in linked list and false otherwise.

bool search(Node *head, int x)

For example, if the key to be searched is 15 and linked list is 14->21->11->30->10, then function should return false. If key to be searched is 14, then the function should return true.
Iterative Solution: 

1) Initialize a node pointer, current = head.
2) Do following while current is not NULL
    a) current->key is equal to the key being searched return true.
    b) current = current->next
3) Return false 

Following is iterative implementation of above algorithm to search a given key.

C




// Iterative C program to search an 
// element in linked list
#include<stdio.h>
#include<stdlib.h>
#include<stdbool.h>
  
// Link list node 
struct Node
{
    int key;
    struct Node* next;
};
  
/* Given a reference (pointer to pointer) to 
   the head of a list and an int, push a new 
   node on the front of the list. */
void push(struct Node** head_ref, 
          int new_key)
{
    // Allocate node
    struct Node* new_node =
           (struct Node*) malloc(sizeof(struct Node));
  
    // Put in the key
    new_node->key = new_key;
  
    // Link the old list off the new node
    new_node->next = (*head_ref);
  
    // Move the head to point to the new node 
    (*head_ref) = new_node;
}
  
// Checks whether the value x is present
// in linked list 
bool search(struct Node* head, int x)
{
    // Initialize current
    struct Node* current = head;  
    while (current != NULL)
    {
        if (current->key == x)
            return true;
        current = current->next;
    }
    return false;
}
  
// Driver code
int main()
{
    // Start with the empty list 
    struct Node* head = NULL;
    int x = 21;
  
    // Use push() to construct list
    // 14->21->11->30->10 
    push(&head, 10);
    push(&head, 30);
    push(&head, 11);
    push(&head, 21);
    push(&head, 14);
  
    search(head, 21)? printf("Yes") : printf("No");
    return 0;
}

Output: 

Yes

Recursive Solution:

bool search(head, x)
1) If head is NULL, return false.
2) If head's key is same as x, return true;
3) Else return search(head->next, x) 

Following is the recursive implementation of the above algorithm to search a given key.

C




// Recursive C program to search an 
// element in linked list
#include<stdio.h>
#include<stdlib.h>
#include<stdbool.h>
// Link list node
struct Node
{
    int key;
    struct Node* next;
};
  
/* Given a reference (pointer to pointer) to 
   the head of a list and an int, push a new
   node on the front of the list. */
void push(struct Node** head_ref, 
          int new_key)
{
    // Allocate node
    struct Node* new_node =
           (struct Node*) malloc(sizeof(struct Node));
  
    // Put in the key
    new_node->key = new_key;
  
    // Link the old list off the new node
    new_node->next = (*head_ref);
  
    // Move the head to point to the new node
    (*head_ref) = new_node;
}
  
// Checks whether the value x is present
// in linked list
bool search(struct Node* head, int x)
{
    // Base case
    if (head == NULL)
        return false;
      
    // If key is present in current 
    // node, return true
    if (head->key == x)
        return true;
  
    // Recur for remaining list
    return search(head->next, x);
}
  
// Driver code
int main()
{
    // Start with the empty list 
    struct Node* head = NULL;
    int x = 21;
  
    // Use push() to construct list
    // 14->21->11->30->10 
    push(&head, 10);
    push(&head, 30);
    push(&head, 11);
    push(&head, 21);
    push(&head, 14);
  
    search(head, 21)? printf("Yes") : printf("No");
    return 0;
}

Output:

Yes

Please refer complete article on Search an element in a Linked List (Iterative and Recursive) for more details!


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!