Javascript Program For Searching An Element In A Linked List

Last Updated : 09 Sep, 2024

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.

// 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

// Head of list
let head;

// Inserts a new node at the front of the list
function push(new_data) {
    // Allocate new node and putting data
    let 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) {
    // Initialize current
    let current = head;

    while (current != null) {
        if (current.data == x)

            // Data found
            return true;
        current = current.next;
    }

    // Data not found
    return false;
}

// Driver code 

// Start with the empty list
// Use push() to construct list
// 14->21->11->30->10
push(10);
push(30);
push(11);
push(21);
push(14);

if (search(head, 21))
    console.log("Yes");
else
    console.log("No");
// This code contributed by aashish1995

Output

Yes

Complexity Analysis:

Time Complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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.

// 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
// Head of list
let head;

// Inserts a new node at the front 
// of the list
function push(new_data) {
    // Allocate new node and putting data
    let 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 code

// Start with the empty list
// Use push() to construct list 
// 14->21->11->30->10
push(10);
push(30);
push(11);
push(21);
push(14);

if (search(head, 21))
    console.log("Yes");
else
    console.log("No");

// This code contributed by gauravrajput1

Output

Yes

Complexity Analysis:

Time Complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(n), for recursive stack where n represents the length of the given linked list.

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