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;}if(n<0){return  head;}//n是合理的    ListNode*  p1=head;ListNode*  p2=head;for(int i=n-1;i>0;--i){        p2=p2->next;}ListNode*  pPre=NULL; //引入这个变量,因为12-2的错误while(p2->next){   pPre=p1;       p2=p2->next;   p1=p1->next;}if(pPre==NULL){head=p1->next;delete p1;}else{ListNode*  pDelete=pPre->next;pPre->next=pDelete->next;delete pDelete;}   return  head;            }};