**Compare two linked lists**

### Problem Statement :

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. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lists are not equal. Return 0. Function Description: Complete the compare_lists function in the editor below. compare_lists has the following parameters: 1. SinglyLinkedListNode llist1: a reference to the head of a list 2. SinglyLinkedListNode llist2: a reference to the head of a list Returns: int: return 1 if the lists are equal, or 0 otherwise Input Format: The first line contains an integer t, the number of test cases. Each of the test cases has the following format: The first line contains an integer n, the number of nodes in the first linked list. Each of the next n lines contains an integer, each a value for a data attribute. The next line contains an integer m, the number of nodes in the second linked list. Each of the next m lines contains an integer, each a value for a data attribute. Constraints: 1. 1<=t<=10 2. 1<=n,m<=1000 3. 1<=list1[i], list2[i]<=1000

### Solution :

` ````
Solution in C :
In C:
//the following function is all that is needed to complete the
//challenge in hackerrank platform.
bool compare_lists(SinglyLinkedListNode* head1, SinglyLinkedListNode* head2) {
struct SinglyLinkedListNode *t1;
struct SinglyLinkedListNode *t2;
t1=head1;t2 = head2;
if(t1==NULL && t2==NULL)
return 1;
if(t1 != NULL && t2 == NULL)
return 0;
if(t1 == NULL && t2 != NULL)
return 0;
else
{
while(t1->next != NULL && t2->next != NULL)
{
if(t1->data == t2->data)
{
t1 = t1->next;
t2 = t2->next;
}
else return 0;
}
if(t1->next == NULL && t2->next == NULL)
return 1;
else return 0;
}
}
In C++:
//the following function is all that is needed to complete the
//challenge in hackerrank platform.
int CompareLists(Node *headA, Node* headB)
{
// This is a "method-only" submission.
// You only need to complete this method
int flag=0;
while(headA != NULL && headB != NULL)
{
if(headA->data!=headB->data)
{
flag=1;
break;
}
headA = headA->next;
headB = headB->next;
}
if(flag==1 || headA!=NULL || headB!=NULL)
return 0;
else
return 1;
}
In Java:
//the following method is all that is needed to complete the
//challenge in hackerrank platform.
static boolean compareLists(SinglyLinkedListNode head1, SinglyLinkedListNode head2) {
if (head1 == null && head2 == null) {
return true;
} else if (head1 == null || head2 == null) {
return false;
} else if (head1.data != head2.data) {
return false;
}
return compareLists(head1.next,head2.next);
}
In Python 3:
//the following method is all that is needed to complete the
//challenge in hackerrank platform.
def CompareLists(headA, headB):
if (headA==None) | (headB==None):
if (headA==None) & (headB==None):
return 1
else:
return 0
else:
while (headA!=None) & (headB!=None):
if (headA.data == headB.data):
headA = headA.next
headB = headB.next
else:
return 0
if (headA == None) & (headB == None):
return 1
else:
return 0
```

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