Remove Nth Node From End of List

Given a linked list, remove the nth 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.

/* *
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
  */
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head,  int n) 
    {
         if(head==NULL)  return NULL;
        
        ListNode* p1=head;
        ListNode* p2=head;
         int count= 1;
         while(count<=n) 
        {
            count++;
            p2=p2->next;
        }
         if(p2==NULL)
             return head->next;
            
         while(p2->next!=NULL)
        {
            p1=p1->next;
            p2=p2->next;
        }
        
        p1->next=p1->next->next;
        
         return head;
    }
};