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Data Structures

## Search an Element

Given a 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

2) Initialize a node pointer, current = head.

3) Do following while current is not NULL

a) current->key is equal to the key being searched return true.

b) current = current->next

4) Return false


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

// Iterative C program to search an element in linked list
#include<stdio.h>
#include<stdlib.h>

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 */

/* move the head to point to the new node */
}

/* Counts no. of nodes in linked list */
bool search(struct node* head, int x)
{
struct node* current = head;  // Initialize current
while (current != NULL)
{
if (current->key == x)
return true;
current = current->next;
}
return false;
}

/* Driver program to test count function*/
int main()
{
int x = 21;

/* Use push() to construct below list
14->21->11->30->10  */

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;
2) Else return search(head->next, x) 
Following is recursive C implementation of above algorithm to search a given key.

// Recursive C program to search an element in linked list
#include<stdio.h>
#include<stdlib.h>

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 */

/* move the head to point to the new node */
}

/* Counts no. of nodes in linked list */
bool search(struct node* head, int x)
{
// Base case
return false;

// If key is present in current node, return true
return true;

// Recur for remaining list
}

/* Driver program to test count function*/
int main()
{
int x = 21;

/* Use push() to construct below list
14->21->11->30->10  */

return 0;
}


Output:

Yes