Middle of a linked list

Method1 (Uses one slow pointer and one fast pointer)
// The slow pointer is advanced only by one node
// and the fast pointer is advanced by two nodes!



typedef struct node
{
  int value;
  struct node *next;
  struct node *prev;
}mynode ;



void getTheMiddle(mynode *head)
{
  mynode *p = head;
  mynode *q = head;

  if(q!=NULL)
  {
       while((q->next)!=NULL && (q->next->next)!=NULL)
       {
          p=(p!=(mynode *)NULL?p->next:(mynode *)NULL);
          q=(q!=(mynode *)NULL?q->next:(mynode *)NULL);
          q=(q!=(mynode *)NULL?q->next:(mynode *)NULL);
       }
       printf("The middle element is [%d]",p->value);
  }
}









Here p moves one step, where as q moves two steps, when q reaches end, p will be at the middle of the linked list.

Method2(Uses a counter)

Function CAll
   middle = getTheMiddle(head);
Function definition

// Function to get to the middle of the LL
mynode *getTheMiddle(mynode *head)
{
  mynode *middle = (mynode *)NULL;
  int i;

  for(i=1; head!=(mynode *)NULL; head=head->next,i++)
  {
     if(i==1)
        middle=head;
     else if ((i%2)==1)
        middle=middle->next;
   }
 
   return middle;
}