Remove Nth Node From End of List

Link: http://oj.leetcode.com/problems/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.

 

//using two pointer, one pointer is n steps ahead of another
    public ListNode removeNthFromEnd(ListNode head, int n) {
      if (head == null||head.next==null)
            return null;
        ListNode p1 = head;
        ListNode p2 = head;
        ListNode result = head;
        //n--;
        while(n-->0){
            p1 = p1.next;
        }
        //special case:
        //input: {1,2} 2
        //in this case, we should delete the head;
        if(p1==null){
            return head.next;
        }
        while (p1.next!=null){
            p1 = p1.next;
            p2 = p2.next;
        }
        p2.next= p2.next.next;
        return result;
    }