给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/remove-nth-node-from-end-of-list
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

示例 1：

输入：head = [1,2,3,4,5], n = 2
输出：[1,2,3,5]

示例 2：

输入：head = [1], n = 1
输出：[]

示例 3：

输入：head = [1,2], n = 1
输出：[1]

思路

• 常规的链表寻找特殊节点，两次遍历利用手动记录的index取操作链表
• 利用快慢指针寻找规律，当next为null时就可以找到该节点

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} n
 * @return {ListNode}
 */
var removeNthFromEnd1 = function (head, n) {

    let count = 1,
        next = head.next;

    while (next) {
        count++
        next = next.next
    }

    let index = count - n + 1

    let result = new ListNode(0, head);
    count = 1
    let tmp = result;

    while (tmp) {
        if (count++ == index) {
            tmp.next = tmp.next.next;
        } else {
            tmp = tmp.next;
        }
    }

    return result.next;
};




var removeNthFromEnd2 = function (head, n) {
    let star = new ListNode(0, head),
        end = new ListNode(0, head);
    
    let result1 = star,
        result2 = end;
    
    let count = 1;
    
    while (result2.next) {
        if (count > n) {
            result1 = result1.next
        }
        result2 = result2.next
        count++
    }

    result1.next = result1.next.next

    return star.next;
};