删除链表中倒数第n个节点-LintCode

给定一个链表，删除链表中倒数第n个节点，返回链表的头节点。
注意事项：
链表中的节点个数大于等于n
样例：
给出链表1->2->3->4->5->null和 n = 2.
删除倒数第二个节点之后，这个链表将变成1->2->3->5->null.
挑战 ：
O(n)时间复杂度

#ifndef C174_H#define C174_H#include<iostream>using namespace std;class ListNode{public:    int val;    ListNode *next;    ListNode(int val)    {        this->val = val;        this->next = NULL;    }};class Solution {public:    /**    * @param head: The first node of linked list.    * @param n: An integer.    * @return: The head of linked list.    */    ListNode *removeNthFromEnd(ListNode *head, int n) {        // write your code here        if (head == NULL)            return NULL;        int len = 0;        ListNode *l = head;        while (l != NULL)        {            len++;            l = l->next;        }        if (len < n||n<=0)            return NULL;        ListNode *node = new ListNode(-1);        node->next = head;        ListNode *p = node, *q = head;        int count = len - n;        while (count != 0)        {            p = p->next;            q = q->next;            count--;        }        p->next = q->next;        q->next = NULL;        return node->next;    }};#endif