Write a C 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

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

Java Programming:

// Iterative Java program to search an element
// in linked list
 
//Node class
class Node
{
    int data;
    Node next;
    Node(int d)
    {
        data = d;
        next = null;
    }
}
 
//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 function to test the above functions
    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);
 
        if (llist.search(llist.head, 21))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
// This code is contributed by Pratik Agarwal
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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)

Java Programming:

// Recursive Java program to search an element
// in linked list
 
 
// Node class
class Node
{
    int data;
    Node next;
    Node(int d)
    {
        data = d;
        next = null;
    }
}
 
// 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 function to test the above functions
    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);
 
        if (llist.search(llist.head, 21))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
// This code is contributed by Pratik Agarwal
[ad type=”banner”]

Output:

Yes

Categorized in:

CodingJAVALinked ListSingly Linked List

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