Search an element in a Linked List (Iterative and Recursive)

Given a linked list and a key ‘X‘ in, the task is to check if X is present in the linked list or not. 

Examples:

Input: 14->21->11->30->10, X = 14
Output: Yes
Explanation: 14 is present in the linked list.

Input: 6->21->17->30->10->8, X = 13
Output: No

Using O(N) Extra Space.

The Approach:

      In this approach we use vector where we store all the nodes value in the vector and then check whether there is key present in vector then it will return 1.

C++

#include <bits/stdc++.h>
using namespace std;

/* Link list node */
class Node {
public:
    int key;
    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(Node** head_ref, int new_key)
{
    /* allocate node */
    Node* new_node = new Node();

    /* put in the key */
    new_node->key = new_key;

    /* link the old list of the new node */
    new_node->next = (*head_ref);

    /* move the head to point to the new node */
    (*head_ref) = new_node;
}

int main() {
     /* Start with the empty list */
    Node* head = NULL;
    int x = 21;

    /* Use push() to construct below list
    14->21->11->30->10 */
    push(&head, 10);
    push(&head, 30);
    push(&head, 11);
    push(&head, 21);
    push(&head, 14);
    vector<int>v;
    //we donot use given data
    Node* temp=head;
    while(temp!=NULL){
     v.push_back(temp->key);
     temp=temp->next;
    }
    // we use iterator to find.
    vector<int>::iterator it;
    find(v.begin(),v.end(),x);
    if(it==v.end()){
      cout<<"NO"<<endl;
    }else{
     cout<<"YES"<<endl;
    }
    return 0;
}

Java

import java.util.*;

class Node {
  int key;
  Node next;

  Node(int key) {
    this.key = key;
    this.next = null;
  }
}

class Main
{

  // 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.
  static void push(Node[] head_ref, int new_key) {
    Node new_node = new Node(new_key);
    new_node.next = head_ref[0];
    head_ref[0] = new_node;
  }

  public static void main(String[] args) {
    Node[] head = new Node[1];
    int x = 21;

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

    // Create a vector of integers from the linked list and search for the value x in the vector using an iterator.
    Vector<Integer> v = new Vector<Integer>();
    Node temp = head[0];
    while (temp != null) {
      v.add(temp.key);
      temp = temp.next;
    }

    int it = v.indexOf(x);
    if (it == -1) {
      System.out.println("NO");
    } else {
      System.out.println("YES");
    }
  }
}

Python3

# Class definition for Node
class Node:
    # Initialize the node with a key
    def __init__(self, key):
        self.key = key
        self.next = None

# Class definition for Linked List
class LinkedList:
    # Initialize the linked list with a head node
    def __init__(self):
        self.head = None

    # Add a new node with key "new_key" at the beginning of the linked list
    def push(self, new_key):
        new_node = Node(new_key)
        new_node.next = self.head
        self.head = new_node

# Create a linked list object
llist = LinkedList()

# Add new nodes to the linked list
llist.push(10)
llist.push(30)
llist.push(11)
llist.push(21)
llist.push(14)

# Key to search for in the linked list
x = 21

# Create a temp variable to traverse the linked list
temp = llist.head

# List to store the keys in the linked list
v = []

# Traverse the linked list and store the keys in the list "v"
while(temp):
    v.append(temp.key)
    temp = temp.next

# Check if "x" is in the list "v"
if x in v:
    print("YES")
else:
    print("NO")

C#

using System;
using System.Collections.Generic;
using System.Linq;

/* Link list node */
class Node {
    public int key;
    public Node next;
};

class Program {
    /* 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. */
    static void push(ref Node head_ref, int new_key)
    {
        /* allocate node */
        Node new_node = new Node();

        /* put in the key */
        new_node.key = new_key;

        /* link the old list of the new node */
        new_node.next = head_ref;

        /* move the head to point to the new node */
        head_ref = new_node;
    }

    static void Main(string[] args)
    {
        /* Start with the empty list */
        Node head = null;
        int x = 21;

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

        List<int> v = new List<int>();
        Node temp = head;
        while (temp != null) {
            v.Add(temp.key);
            temp = temp.next;
        }

        /* Find whether x is present in the list or not */
        if (v.Contains(x)) {
            Console.WriteLine("YES");
        }
        else {
            Console.WriteLine("NO");
        }
    }
}
// This code is contributed by sarojmcy2e

Javascript

// Class definition for Node
class Node {
// Initialize the node with a key
constructor(key) {
this.key = key;
this.next = null;
}
}

// Class definition for Linked List
class LinkedList {
// Initialize the linked list with a head node
constructor() {
this.head = null;
}

// Add a new node with key "new_key" at the beginning of the linked list
push(new_key) {
let new_node = new Node(new_key);
new_node.next = this.head;
this.head = new_node;
}
}

// Create a linked list object
let llist = new LinkedList();

// Add new nodes to the linked list
llist.push(10);
llist.push(30);
llist.push(11);
llist.push(21);
llist.push(14);

// Key to search for in the linked list
let x = 22;

// Create a temp variable to traverse the linked list
let temp = llist.head;

// Array to store the keys in the linked list
let v = [];

// Traverse the linked list and store the keys in the array "v"
while (temp) {
v.push(temp.key);
temp = temp.next;
}

// Check if "x" is in the array "v"
if (v.includes(x)) {
console.log("YES");
} else {
console.log("NO");
}

Output

YES

Time Complexity: O(N), to traverse linked list.
Auxiliary Space: O(N),to store the values.

Search an element in a Linked List (Iterative Approach): 

Follow the below steps to solve the problem:

• Initialize a node pointer, current = head.
• Do following while current is not NULL
  •  If the current value (i.e., current->key) is equal to the key being searched return true.
  • Otherwise, move to the next node (current = current->next).
• If the key is not found, return false 

Below is the implementation of the above approach.

C++

// Iterative C++ program to search
// an element in linked list
#include <bits/stdc++.h>
using namespace std;

/* Link list node */
class Node {
public:
    int key;
    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(Node** head_ref, int new_key)
{
    /* allocate node */
    Node* new_node = new Node();

    /* put in the key */
    new_node->key = new_key;

    /* link the old list of 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(Node* head, int x)
{
    Node* current = head; // Initialize current
    while (current != NULL) {
        if (current->key == x)
            return true;
        current = current->next;
    }
    return false;
}

/* Driver code*/
int main()
{
    /* Start with the empty list */
    Node* head = NULL;
    int x = 21;

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

      // Function call
    search(head, x) ? cout << "Yes" : cout << "No";
    return 0;
}

// This is code is contributed by rathbhupendra

C

// Iterative C program to search an element
// in the linked list
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.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 of 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)
{
    struct Node* current = head; // Initialize current
    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 below list
     14->21->11->30->10  */
    push(&head, 10);
    push(&head, 30);
    push(&head, 11);
    push(&head, 21);
    push(&head, 14);

      // Function call
    search(head, x) ? printf("Yes") : printf("No");
    return 0;
}

Java

// Iterative Java program to search an element
// in linked list

// Linked list class
class LinkedList {
    Node head; // Head of list

    // Inserts a new node at the front of the list
    public void push(int new_data)
    {
        // Allocate new node and putting data
        Node new_node = new Node(new_data);

        // Make next of new node as head
        new_node.next = head;

        // Move the head to point to new Node
        head = new_node;
    }

    // Checks whether the value x is present in linked list
    public boolean search(Node head, int x)
    {
        Node current = head; // Initialize current
        while (current != null) {
            if (current.data == x)
                return true; // data found
            current = current.next;
        }
        return false; // data not found
    }

    // Driver code
    public static void main(String args[])
    {

        // Start with the empty list
        LinkedList llist = new LinkedList();
        int x = 21;
        /*Use push() to construct below list
        14->21->11->30->10  */
        llist.push(10);
        llist.push(30);
        llist.push(11);
        llist.push(21);
        llist.push(14);

          // Function call
        if (llist.search(llist.head, x))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// Node class
class Node {
    int data;
    Node next;
    Node(int d)
    {
        data = d;
        next = null;
    }
}

// This code is contributed by Pratik Agarwal

Python3

# Iterative Python3 program to search an element
# in linked list

# Node class


class Node:

    # Function to initialise the node object
    def __init__(self, data):
        self.data = data  # Assign data
        self.next = None  # Initialize next as null

# Linked List class


class LinkedList:
    def __init__(self):
        self.head = None  # Initialize head as None

    # This function insert a new node at the
    # beginning of the linked list
    def push(self, new_data):

        # Create a new Node
        new_node = Node(new_data)

        # 3. Make next of new Node as head
        new_node.next = self.head

        # 4. Move the head to point to new Node
        self.head = new_node

    # This Function checks whether the value
    # x present in the linked list
    def search(self, x):

        # Initialize current to head
        current = self.head

        # loop till current not equal to None
        while current != None:
            if current.data == x:
                return True  # data found

            current = current.next

        return False  # Data Not found


# Driver code
if __name__ == '__main__':

    # Start with the empty list
    llist = LinkedList()
    x = 21

    ''' Use push() to construct below list
        14->21->11->30->10 '''
    llist.push(10)
    llist.push(30)
    llist.push(11)
    llist.push(21)
    llist.push(14)

       # Function call
    if llist.search(x):
        print("Yes")
    else:
        print("No")

# This code is contributed by Ravi Shankar

C#

// Iterative C# program to search an element
// in linked list
using System;

// Node class
public class Node {
    public int data;
    public Node next;
    public Node(int d)
    {
        data = d;
        next = null;
    }
}

// Linked list class
public class LinkedList {
    Node head; // Head of list

    // Inserts a new node at the front of the list
    public void push(int new_data)
    {
        // Allocate new node and putting data
        Node new_node = new Node(new_data);

        // Make next of new node as head
        new_node.next = head;

        // Move the head to point to new Node
        head = new_node;
    }

    // Checks whether the value x is present in linked list
    public bool search(Node head, int x)
    {
        Node current = head; // Initialize current
        while (current != null) {
            if (current.data == x)
                return true; // data found
            current = current.next;
        }
        return false; // data not found
    }

    // Driver code
    public static void Main(String[] args)
    {

        // Start with the empty list
        LinkedList llist = new LinkedList();

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

        int x = 21;
        // Function call
        if (llist.search(llist.head, x))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

// This code contributed by Rajput-Ji

Javascript

// Iterative javascript program
// to search an element
// in linked list

//Node class
class Node {
    constructor(d) {
        this.data = d;
        this.next = null;
    }
}

// Linked list class

    var head; // Head of list

    // Inserts a new node at the front of the list
    function push(new_data)
    {
        // Allocate new node and putting data
        var new_node = new Node(new_data);

        // Make next of new node as head
        new_node.next = head;

        // Move the head to point to new Node
        head = new_node;
    }

    // Checks whether the value
    // x is present in linked list
    function search( head , x)
    {
        var current = head; // Initialize current
        while (current != null) {
            if (current.data == x)
                return true; // data found
            current = current.next;
        }
        return false; // data not found
    }

    // Driver function to test
    // the above functions


        // Start with the empty list
        /*
          Use push() to construct below
          list 14->21->11->30->10
         */
        push(10);
        push(30);
        push(11);
        push(21);
        push(14);
        var x = 21;
        if (search(head, x))
            document.write("Yes");
        else
            document.write("No");

// This code contributed by aashish1995

Output

Yes

Time Complexity: O(N), Where N is the number of nodes in the LinkedList
Auxiliary Space: O(1)

Search an element in a Linked List (Recursive Approach): 

Follow the below steps to solve the problem:

• If the head is NULL, return false.
• If the head’s key is the same as X, return true;
• Else recursively search in the next node. 

Below is the recursive implementation of the above algorithm.

C++

// Recursive C++ program to search
// an element in linked list
#include <bits/stdc++.h>
using namespace std;

/* 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 of 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 below list
    14->21->11->30->10 */
    push(&head, 10);
    push(&head, 30);
    push(&head, 11);
    push(&head, 21);
    push(&head, 14);

      // Function call
    search(head, x) ? cout << "Yes" : cout << "No";
    return 0;
}

// This code is contributed by SHUBHAMSINGH10

C

// Recursive C program to search an element in linked list
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.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 of 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 below list
     14->21->11->30->10  */
    push(&head, 10);
    push(&head, 30);
    push(&head, 11);
    push(&head, 21);
    push(&head, 14);

      // Function call
    search(head, x) ? printf("Yes") : printf("No");
    return 0;
}

Java

// Recursive Java program to search an element
// in the linked list

// Linked list class
class LinkedList {
    Node head; // Head of list

    // Inserts a new node at the front of the list
    public void push(int new_data)
    {
        // Allocate new node and putting data
        Node new_node = new Node(new_data);

        // Make next of new node as head
        new_node.next = head;

        // Move the head to point to new Node
        head = new_node;
    }

    // Checks whether the value x is present
    // in linked list
    public boolean search(Node head, int x)
    {
        // Base case
        if (head == null)
            return false;

        // If key is present in current node,
        // return true
        if (head.data == x)
            return true;

        // Recur for remaining list
        return search(head.next, x);
    }

    // Driver code
    public static void main(String args[])
    {
        // Start with the empty list
        LinkedList llist = new LinkedList();

        /* Use push() to construct below list
           14->21->11->30->10  */
        llist.push(10);
        llist.push(30);
        llist.push(11);
        llist.push(21);
        llist.push(14);
        int x = 21;
          // Function call
        if (llist.search(llist.head, x))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// Node class
class Node {
    int data;
    Node next;
    Node(int d)
    {
        data = d;
        next = null;
    }
}
// This code is contributed by Pratik Agarwal

Python3

# Recursive Python program to
# search an element in linked list

# Node class


class Node:

    # Function to initialise
    # the node object
    def __init__(self, data):
        self.data = data  # Assign data
        self.next = None  # Initialize next as null


class LinkedList:

    def __init__(self):
        self.head = None  # Initialize head as None

    # This function insert a new node at
    # the beginning of the linked list
    def push(self, new_data):

        # Create a new Node
        new_node = Node(new_data)

        # Make next of new Node as head
        new_node.next = self.head

        # Move the head to
        # point to new Node
        self.head = new_node

    # Checks whether the value key
    # is present in linked list

    def search(self, li, key):

        # Base case
        if(not li):
            return False

        # If key is present in
        # current node, return true
        if(li.data == key):
            return True

        # Recur for remaining list
        return self.search(li.next, key)


# Driver Code
if __name__ == '__main__':

    li = LinkedList()

    li.push(10)
    li.push(30)
    li.push(11)
    li.push(21)
    li.push(14)

    x = 21

    # Function call
    if li.search(li.head, x):
        print("Yes")
    else:
        print("No")

# This code is contributed
# by Manoj Sharma

C#

// Recursive C# program to search
// an element in linked list
using System;

// Node class
public class Node {
    public int data;
    public Node next;
    public Node(int d)
    {
        data = d;
        next = null;
    }
}

// Linked list class
public class LinkedList {
    Node head; // Head of list

    // Inserts a new node at the front of the list
    public void push(int new_data)
    {
        // Allocate new node and putting data
        Node new_node = new Node(new_data);

        // Make next of new node as head
        new_node.next = head;

        // Move the head to point to new Node
        head = new_node;
    }

    // Checks whether the value x is present
    // in linked list
    public bool search(Node head, int x)
    {
        // Base case
        if (head == null)
            return false;

        // If key is present in current node,
        // return true
        if (head.data == x)
            return true;

        // Recur for remaining list
        return search(head.next, x);
    }

    // Driver code
    public static void Main()
    {
        // Start with the empty list
        LinkedList llist = new LinkedList();

        /* Use push() to construct below list
        14->21->11->30->10 */
        llist.push(10);
        llist.push(30);
        llist.push(11);
        llist.push(21);
        llist.push(14);
        int x = 21;
        // Function call
        if (llist.search(llist.head, x))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

// This code is contributed by PrinciRaj1992

Javascript

// Recursive javascript program to search an element
// in linked list

// Node class
 class Node {
        constructor(val) {
            this.data = val;
            this.next = null;
        }
    }

// Linked list class
var head; // Head of list

    // Inserts a new node at the front of the list
     function push(new_data) {
        // Allocate new node and putting data
var new_node = new Node(new_data);

        // Make next of new node as head
        new_node.next = head;

        // Move the head to point to new Node
        head = new_node;
    }

    // Checks whether the value x is present
    // in linked list
     function search(head , x) {
        // Base case
        if (head == null)
            return false;

        // If key is present in current node,
        // return true
        if (head.data == x)
            return true;

        // Recur for remaining list
        return search(head.next, x);
    }

    // Driver function to test the above functions

        // Start with the empty list


        /*
         * Use push() to construct below list 14->21->11->30->10
         */
        push(10);
        push(30);
        push(11);
        push(21);
        push(14);
        var x = 21;
        if (search(head, 21))
            document.write("Yes");
        else
            document.write("No");

// This code contributed by gauravrajput1

Output

Yes

Time Complexity: O(N)
Auxiliary Space: O(N), Stack space used by recursive calls

This article is contributed by Ravi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above