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||n==0)return head;
        ListNode *p, *np;
        p = head;
        np = head;
        while(np)
        {
            np = np->next;
            n--;
            if(n==0)
                break;
        }
        if(!np&&n==0)
            return p->next;
        if(!np&&n!=0)
            return NULL;
        np = np->next;
        while(np)
        {
            p  = p->next;
            np = np->next;
        }
        p->next = p->next->next;
        return head;
    }
};

代码略长, 边界状态慎重