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 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
{ // Head of list
Node head;
// 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)
{
// Initialize current
Node current = head;
while (current != null )
{
// Data found
if (current.data == x)
return true ;
current = current.next;
}
// Data not found
return false ;
}
// Driver code
public static void main(String args[])
{
// Start with the empty list
LinkedList llist = new LinkedList();
// Use push() to construct 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 |
Output:
Yes
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 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
{ // Head of list
Node head;
// 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 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 |
Output:
Yes
Time Complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(n), for recursive call 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!