Python Program To Check If Two Linked Lists Are Identical

Two Linked Lists are identical when they have the same data and the arrangement of data is also the same. For example, Linked lists a (1->2->3) and b(1->2->3) are identical. . Write a function to check if the given two linked lists are identical. 

Method 1 (Iterative):
To identify if two lists are identical, we need to traverse both lists simultaneously, and while traversing we need to compare data.

Python3

# An iterative Java program to check if
# two linked lists are identical or not

# Linked list Node
class Node:
    def __init__(self, d):
        self.data = d
        self.next = None

class LinkedList:
    def __init__(self):

        # Head of list
        self.head = None

    # Returns true if linked lists a
    # and b are identical, otherwise false
    def areIdentical(self, listb):
        a = self.head
        b = listb.head

        while (a != None and b != None):
            if (a.data != b.data):
                return False

            # If we reach here, then a and b
            # are not null and their data is
            # same, so move to next nodes
            # in both lists
            a = a.next
            b = b.next

        # If linked lists are identical,
        # then 'a' and 'b' must be null
        # at this point.
        return (a == None and b == None)

    # UTILITY FUNCTIONS TO TEST fun1()
    # and fun2()
    # Given a reference (pointer to pointer)
    # to the head of a list and an int, push
    # a new node on the front of the list.
    def push(self, new_data):

        # 1 & 2: Allocate the Node &
        # Put in the data
        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

# Driver Code
llist1 = LinkedList()
llist2 = LinkedList()

# The constructed linked lists are :
# llist1: 3->2->1
# llist2: 3->2->1
llist1.push(1)
llist1.push(2)
llist1.push(3)
llist2.push(1)
llist2.push(2)
llist2.push(3)

if (llist1.areIdentical(llist2) == True):
    print("Identical ")
else:
    print("Not identical ")
# This code is contributed by Prerna Saini

Output:

Identical

Time Complexity: O(n)

Method 2 (Recursive):
Recursive solution code is much cleaner than iterative code. You probably wouldn’t want to use the recursive version for production code, however, because it will use stack space which is proportional to the length of the lists.

Python3

# A recursive Python3 function to check
# if two linked lists are identical
# or not
def areIdentical(a, b):

    # If both lists are empty
    if (a == None and b == None):
        return True

    # If both lists are not empty,
    # then data of current nodes must
    # match, and same should be recursively
    # true for rest of the nodes.
    if (a != None and b != None):
        return ((a.data == b.data) and
                 areIdentical(a.next, b.next))

    # If we reach here, then one of the lists
    # is empty and other is not
    return False
# This code is contributed by Princi Singh

Time Complexity: O(n) for both iterative and recursive versions. n is the length of the smaller list among a and b.

Auxiliary Space: O(n) for call stack because using recursion

Approach 3: Hash Table:

• Another approach is to compare two linked lists is a hash table or a set. The idea is to traverse the first linked list and insert each node’s value into the hash table or set. 
• Then, traverse the second linked list and check if each node’s value exists in the hash table or set. If a node’s value is not found in the hash table or set, then the linked lists are not identical.

Here’s a simple Python Code that uses a hash table (unordered_set) to compare two linked lists:

Python3

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def areIdentical(head1, head2):
    s = set()
    while head1 != None:
        s.add(head1.data)
        head1 = head1.next
    while head2 != None:
        if head2.data not in s:
            return False
        head2 = head2.next
    return True

def push(head, data):
    new_node = Node(data)
    new_node.next = head
    head = new_node
    return head

if __name__ == '__main__':
    head1 = None
    head2 = None
    head1 = push(head1, 1)
    head1 = push(head1, 2)
    head1 = push(head1, 3)
    head2 = push(head2, 1)
    head2 = push(head2, 2)
    head2 = push(head2, 3)
    if areIdentical(head1, head2):
        print("Identical")
    else:
        print("Not identical")

Output

Identical

Time Complexity: O(n) where n is the length of the smaller list among a and b.

Space complexity: O(n), where n is the number of elements stored in the table. 

Please refer complete article on Identical Linked Lists for more details!