# Description

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.

The original problem is here.

The original code is here.

# My Solution

I solve this problem in C++, as below:

``````/*
*Remove Nth Node from End of List
*Author: shuaijiang
*Email: zhaoshuaijiang8@gmail.com
*/
#include<iostream>
#include<stdlib.h>
using namespace std;

/**
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
return NULL;
if(head->next == NULL && n == 1)
return NULL;
ListNode *ptr, *ptr_nth, *ptr_before;
int count  = 0;
while(ptr->next != NULL){
if(count > n){
ptr_before = ptr_nth;
ptr_nth = ptr_nth->next;
}
count ++;
ptr = ptr->next;
}
if(ptr_before == ptr_nth)