### 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:

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.

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.

{
// This is a "method-only" submission.
// You only need to complete this method
int flag=0;

{
{
flag=1;
break;
}
}

return 0;
else
return 1;

}

In Java:

//the following method is all that is needed to complete the
//challenge in hackerrank platform.

return true;
return false;
return false;
}

}

In Python 3:

//the following method is all that is needed to complete the
//challenge in hackerrank platform.

return 1
else:
return 0
else:
else:
return 0
return 1
else:
return 0```
```

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