leetcode: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) {}
 * };
 */
#include <iostream>
using namespace std;

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
    ListNode* end_p,* p,*pre_p;

    end_p=p=pre_p=head;


    if(p==NULL) return head;
    else if(p->next == NULL )
    {
        delete(p);
        head=NULL;
        return head;
    }

    n--;
    while(n)
    {
        end_p = end_p->next;
        n--;
    }


    while(end_p->next!=NULL)
    {
        pre_p=p;
        end_p=end_p->next;
        p=p->next;
    }


    if(pre_p==p) head = p->next;
    else pre_p->next=p->next;
    delete(p);


    return head;
    }
};