Compare two linked lists - Hacker Rank Solution

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. The lists are equal only if they have the same number of nodes and corresponding nodes contain the same data. Either head pointer given may be null meaning that the corresponding list is empty.

Input Format
You have to complete the int CompareLists(Node* headA, Node* headB) method which takes two arguments - the heads of the two linked lists to compare. You should NOT read any input from stdin/console.
Output Format
Compare the two linked lists and return 1 if the lists are equal. Otherwise, return 0. Do NOT print anything to stdout/console.
Sample Input
NULL, 1 --> NULL
1 --> 2 --> NULL, 1 --> 2 --> NULL
Sample Output

0
1

Explanation
1. We compare an empty list with a list containing 1. They don't match, hence return 0.
2. We have 2 similar lists. Hence return 1.
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 /*
  Compare two linked lists A and B
  Return 1 if they are identical and 0 if they are not.
  Node is defined as
  struct Node
  {
     int data;
     struct Node *next;
  }
*/
int CompareLists(Node *headA, Node* headB)
{
  // This is a "method-only" submission.
  // You only need to complete this method
    while((headA!= NULL) || (headB!= NULL)){ 
     if((headA == NULL) || (headB == NULL)) 
       return 0; 
     if(headA->data != headB->data){ 
       return 0; 
     } 
     headA = headA->next; 
     headB = headB->next; 
   } 
   return 1;
}
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