Python 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.

Python

# Iterative Python program to search 
# an element in linked list

# Node class
class Node:
    
    # Function to initialise the 
    # node object
    def __init__(self, data):
    
        # Assign data
        self.data = data 

        # Initialize next as null
        self.next = None 

# Linked List class
class LinkedList:
    def __init__(self):

        # Initialize head as None
        self.head = None 

    # This function insert a new node at the
    # beginning of the linked list
    def push(self, new_data):
    
        # Create a new Node
        new_node = Node(new_data)

        # 3. Make next of new Node as head
        new_node.next = self.head

        # 4. Move the head to point to new Node
        self.head = new_node

    # This Function checks whether the value
    # x present in the linked list 
    def search(self, x):

        # Initialize current to head
        current = self.head

        # Loop till current not equal to None
        while current != None:
            if current.data == x:

                # Data found
                return True 
            
            current = current.next
        
        # Data Not found
        return False 

# Driver code
if __name__ == '__main__':

    # Start with the empty list
    llist = 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(21):
        print("Yes")
    else:
        print("No")
# This code is contributed by Ravi Shankar

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.

Python

# Recursive Python program to 
# search an element in linked list

# Node class
class Node:
    
    # Function to initialize 
    # the node object
    def __init__(self, data):

        # Assign data
        self.data = data 

        # Initialize next as null
        self.next = None 

class LinkedList:
    
    def __init__(self):

        # Initialize head as None
        self.head = None 

    # This function insert a new node at 
    # the beginning of the linked list
    def push(self, new_data):
    
        # Create a new Node
        new_node = Node(new_data)

        # Make next of new Node as head
        new_node.next = self.head

        # Move the head to 
        # point to new Node
        self.head = new_node
    
    
    # Checks whether the value key 
    # is present in linked list 
    def search(self, li, key):
        
        # Base case
        if(not li):
            return False
        
        # If key is present in 
        # current node, return true
        if(li.data == key):
            return True
        
        # Recur for remaining list
        return self.search(li.next, key)
    
# Driver Code            
if __name__=='__main__':

    li = LinkedList()
    
    li.push(1)
    li.push(2)
    li.push(3)
    li.push(4)
    
    key = 4
    
    if li.search(li.head,key):
        print("Yes")
    else:
        print("No")
# This code is contributed by Manoj Sharma

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!