Find the first intersection of two lists (if the two list are merged at some point)

http://www.geeksforgeeks.org/write-a-function-to-get-the-intersection-point-of-two-linked-lists/
Method 3(Using difference of node counts)
1) Get count of the nodes in first list, let count be c1.
2) Get count of the nodes in second list, let count be c2.
3) Get the difference of counts d = abs(c1 – c2)
4) Now traverse the bigger list from the first node till d nodes so that from here onwards both the lists have equal no of nodes.
5) Then we can traverse both the lists in parallel till we come across a common node. (Note that getting a common node is done by comparing the address of the nodes)

/* function to get the intersection point of two linked   lists head1 and head2 */int getIntesectionNode(struct node* head1, struct node* head2){  int c1 = getCount(head1);  int c2 = getCount(head2);  int d;  if(c1 > c2)  {    d = c1 - c2;    return _getIntesectionNode(d, head1, head2);  }  else  {    d = c2 - c1;    return _getIntesectionNode(d, head2, head1);  }}/* function to get the intersection point of two linked   lists head1 and head2 where head1 has d more nodes than   head2 */int _getIntesectionNode(int d, struct node* head1, struct node* head2){  int i;  struct node* current1 = head1;  struct node* current2 = head2;  for(i = 0; i < d; i++)  {    if(current1 == NULL)    {  return -1; }    current1 = current1->next;  }  while(current1 !=  NULL && current2 != NULL)  {    if(current1 == current2)      return current1->data;    current1= current1->next;    current2= current2->next;  }  return -1;}/* Takes head pointer of the linked list and   returns the count of nodes in the list */int getCount(struct node* head){  struct node* current = head;  int count = 0;  while (current != NULL)  {    count++;    current = current->next;  }  return count;}