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