Delete nth Node from the last of Link List

Given a linked list, remove the n^th node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

   ListNode *removeNthFromEnd(ListNode *head, int n) {

        if(n <= 0  || !head)

                return head;

        // take slow pointer and fast pointer

        ListNode *slow = NULL;

        ListNode *fast = NULL;

        // move fast pointer to n nodes

        while(n > 0)

        {  

            if(!fast)

                fast = head;

            else 

                fast = fast->next;

         

            if(!fast)

                break;

            n--;

        }

        if(!fast)

            return head;

        fast = fast->next;

        // now move slow pointer and fast pointer

        ListNode *prevNode = NULL;

        slow = head;

        while(fast)

        {

            if(!prevNode)

            {

                prevNode = slow;

            }

            else {

                prevNode = prevNode->next;

            }

            fast = fast->next;

        }

        // now need  

        if(!prevNode)

            head =  head->next;

        else

            prevNode->next = prevNode->next->next;

        return head;

}