Java Program For Searching An Element In A Linked List
Last Updated :
15 Jun, 2022
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.
Java
class Node
{
int data;
Node next;
Node( int d)
{
data = d;
next = null ;
}
}
class LinkedList
{
Node head;
public void push( int new_data)
{
Node new_node = new Node(new_data);
new_node.next = head;
head = new_node;
}
public boolean search(Node head, int x)
{
Node current = head;
while (current != null )
{
if (current.data == x)
return true ;
current = current.next;
}
return false ;
}
public static void main(String args[])
{
LinkedList llist = new LinkedList();
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" );
}
}
|
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.
Java
class Node
{
int data;
Node next;
Node( int d)
{
data = d;
next = null ;
}
}
class LinkedList
{
Node head;
public void push( int new_data)
{
Node new_node = new Node(new_data);
new_node.next = head;
head = new_node;
}
public boolean search(Node head, int x)
{
if (head == null )
return false ;
if (head.data == x)
return true ;
return search(head.next, x);
}
public static void main(String args[])
{
LinkedList llist = new LinkedList();
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" );
}
}
|
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!
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