Java Program For Comparing Two Strings Represented As Linked Lists
Last Updated :
23 Jul, 2025
Given two strings, represented as linked lists (every character is a node in a linked list). Write a function compare() that works similar to strcmp(), i.e., it returns 0 if both strings are the same, 1 if the first linked list is lexicographically greater, and -1 if the second string is lexicographically greater.
Examples:
Input: list1 = g->e->e->k->s->a
list2 = g->e->e->k->s->b
Output: -1
Input: list1 = g->e->e->k->s->a
list2 = g->e->e->k->s
Output: 1
Input: list1 = g->e->e->k->s
list2 = g->e->e->k->s
Output: 0
Java
// Java program to compare two strings
// represented as a linked list
// Linked List Class
class LinkedList
{
// Head of list
Node head;
static Node a, b;
// Node Class
static class Node
{
char data;
Node next;
// Constructor to create
// a new node
Node(char d)
{
data = d;
next = null;
}
}
int compare(Node node1,
Node node2)
{
if (node1 == null &&
node2 == null)
{
return 1;
}
while (node1 != null &&
node2 != null &&
node1.data == node2.data)
{
node1 = node1.next;
node2 = node2.next;
}
// if the list are different
// in size
if (node1 != null &&
node2 != null)
{
return (node1.data >
node2.data ? 1 : -1);
}
// if either of the list has
// reached end
if (node1 != null &&
node2 == null)
{
return 1;
}
if (node1 == null &&
node2 != null)
{
return -1;
}
return 0;
}
// Driver code
public static void main(String[] args)
{
LinkedList list = new LinkedList();
Node result = null;
list.a = new Node('g');
list.a.next = new Node('e');
list.a.next.next = new Node('e');
list.a.next.next.next =
new Node('k');
list.a.next.next.next.next =
new Node('s');
list.a.next.next.next.next.next =
new Node('b');
list.b = new Node('g');
list.b.next = new Node('e');
list.b.next.next = new Node('e');
list.b.next.next.next =
new Node('k');
list.b.next.next.next.next =
new Node('s');
list.b.next.next.next.next.next =
new Node('a');
int value;
value = list.compare(a, b);
System.out.println(value);
}
}
// This code is contributed by Mayank Jaiswal
Output:
1
Time Complexity: O(M + N), where M and N represents the length of the given two linked lists.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Compare two strings represented as linked lists for more details!
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