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.
Java
// An iterative Java program to check if two // linked lists are identical or notclass LinkedList{ // Head of list Node head; // Linked list Node class Node { int data; Node next; Node(int d) { data = d; next = null; } } /* Returns true if linked lists a and b are identical, otherwise false */ boolean areIdentical(LinkedList listb) { Node a = this.head, b = listb.head; while (a != null && b != null) { 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 == null && b == null); } // 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. */ void push(int new_data) { /* 1 & 2: Allocate the Node & Put in the data*/ Node new_node = new Node(new_data); // 3. Make next of new Node as head new_node.next = head; // 4. Move the head to point to new Node head = new_node; } // Driver code public static void main(String args[]) { LinkedList llist1 = new LinkedList(); LinkedList llist2 = new 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) System.out.println("Identical "); else System.out.println("Not identical "); }} // This code is contributed by Rajat Mishra |
Output:
Identical
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.
Java
// A recursive Java method to check if two // linked lists are identical or notboolean areIdenticalRecur(Node a, Node b){ // If both lists are empty if (a == null && b == null) 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 != null && b != null) return (a.data == b.data) && areIdenticalRecur(a.next, b.next); // If we reach here, then one of the lists // is empty and other is not return false;}/* Returns true if linked lists a and b are identical, otherwise false */boolean areIdentical(LinkedList listb){ return areIdenticalRecur(this.head, listb.head);} |
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
Please refer complete article on Identical Linked Lists for more details!
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