Intersection point in Y shaped linked list

There are two singly linked lists in a system. By some programming error the end node of one of the linked list got linked into the second list, forming a inverted Y shaped list. Write a program to get the point where two linked lists merge.

Above diagram shows an example with two linked list having 15 as intersection point.

Expected time complexity is O(m + n) where m and n are lengths of two linked lists

Input:

You have to complete the method which takes two arguments as heads of two linked lists.  

Output:
The function should return data value of a node where two linked lists merge.  If linked list do not merge at any point, then it shoudl return -1.

Constraints:
1 <=T<= 50
1 <=N<= 100
1 <=value<= 1000

Example:
Input:
2
2 3 2
10 20
30 40 50
5 10
2 3 0
10 20
30 40 50
First line is number of test cases. Every test case has four lines. First line of every test case contains three numbers, x (number of nodes before merge point in 1st list), y(number of nodes before merge point in 11nd list) and z (number of nodes after merge point).  Next three lines contain x, y and z values.

Output:
5
-1

Solution:-

Here's a simple trick. As the values are always positive, so traverse through the first linked list, and multiply all values by -1.
Now start traversing from the second head. If any negative value is encountered, means that is the merge point. So return negative of that value.

int intersectPoint(Node* head1, Node* head2){
Node *temp = head1;
while(temp){
temp->data = -1 * temp->data;
temp = temp->next;
}
temp = head2;
int ans = -1;
while(temp){
if(temp->data < 0){
ans = -temp->data;
break;
}
temp = temp->next;
}
return ans;
}