删除倒数第 N 个节点

题目描述
给定一个链表，删除链表的倒数第n个节点并返回链表的头指针
例如，
给出的链表为:1->2->3->4->5, n= 2.
删除了链表的倒数第n个节点之后,链表变为1->2->3->5.
备注：
题目保证n一定是有效的
请给出请给出时间复杂度为 O(n) O(n)的算法

示例1
输入
复制
{1,2},2
返回值
复制
{2}

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param head ListNode类 
     * @param n int整型 
     * @return ListNode类
     */
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        // write code here
        ListNode * myHead = new ListNode(0),*slow = myHead, *fast = myHead;
        myHead ->next = head;
        while(n--) fast = fast ->next;
        // 快指针先 走 n 下
        while(fast ->next) slow= slow->next, fast = fast ->next;
        auto will_del = slow ->next;
        slow ->next = slow ->next ->next;
        delete will_del;
        will_del = myHead;
        myHead = myHead ->next;
        delete will_del;
        return myHead;
    }
    
    ListNode * reverse(ListNode* p) {
        ListNode* pre = NULL,*next = NULL;
        while (p) {
            next = p->next,p->next = pre,pre = p ,p=next;
        }
        return pre;
    }
};