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。

示例：

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

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * 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 removeNthFromEnd = function(head, n) {
    let list = new ListNode(0, head)
    let copyList = list
    let stack = []
    while (copyList) {
      stack.push(copyList)
      copyList = copyList.next
    }

    for (let i = 0; i < n; i++) {
      stack.pop()
    }

    let pre = stack[stack.length - 1]
    pre.next = pre.next.next

    return list.next

};

/**
 * 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 removeNthFromEnd = function(head, n) {
    let list = head
    if (!list.next) return null
    let size = 1
    while (list.next) {
      size++
      list = list.next
    }

    if (n === size) return head.next

    let current = head
    let previous = null
    let start = 0
    while (start < size - n) {
      previous = current
      current = current.next
        ++start
    }

    previous.next = current.next

    return head

};

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/