Javascript Program For Comparing Two Strings Represented As Linked Lists
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
Javascript
<script> // Javascript program to compare two // strings represented as a linked list // Linked List Class // head of list var head; var a, b; // Node Class class Node { // Constructor to create a // new node constructor(d) { this .data = d; this .next = null ; } } function compare(node1, 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 var result = null ; a = new Node( 'g' ); a.next = new Node( 'e' ); a.next.next = new Node( 'e' ); a.next.next.next = new Node( 'k' ); a.next.next.next.next = new Node( 's' ); a.next.next.next.next.next = new Node( 'b' ); b = new Node( 'g' ); b.next = new Node( 'e' ); b.next.next = new Node( 'e' ); b.next.next.next = new Node( 'k' ); b.next.next.next.next = new Node( 's' ); b.next.next.next.next.next = new Node( 'a' ); var value; value = compare(a, b); document.write(value); // This code is contributed by gauravrajput1 </script> |
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.
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