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
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Output:
Yes