19. 删除链表的倒数第 N 个结点

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 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]

提示：
链表中结点的数目为 sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz

进阶：你能尝试使用一趟扫描实现吗？

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

javascript

/**
 * @param {ListNode} head
 * @param {number} n
 * @return {ListNode}
 */
var removeNthFromEnd = function (head, n) {
  if (n === 1 && head && !head.next) {
    return null
  }
  let pre = head,
    cur = head,
    i = 0
  for (; pre; pre = pre.next) {
    i++
  }
  if (i === n) {
    cur = cur.next
    return cur
  }
  const k = i - n
  for (let j = 1; j < k; j++) {
    cur = cur.next
  }
  cur.next = n === 1 ? null : cur.next.next
  return head
}

var removeNthFromEnd = function (head, n) {
  let temp = head,
    list = [head],
    i = 1
  while (temp.next) {
    temp = temp.next
    list.push(temp)
    i++
  }
  if (i === 1) {
    return null
  }
  const remove = i - n
  if (remove === 0) {
    return head.next
  }
  const pre = list[remove - 1]
  pre.next = pre.next.next
  return head
}

var removeNthFromEnd = function (head, n) {
  const newNode = new ListNode(0)
  newNode.next = head
  let left = newNode,
    right = newNode,
    i = 0
  while (i <= n) {
    right = right.next
    i++
  }
  while (right) {
    left = left.next
    right = right.next
  }
  if (left && left.next) {
    left.next = left.next.next
  }

  return newNode.next
}

// test
function ListNode(val, next) {
  this.val = val === undefined ? 0 : val
  this.next = next === undefined ? null : next
}

var l1 = new ListNode(1)
var lstNode1 = new ListNode(2)
var lstNode2 = new ListNode(3)
var lstNode3 = new ListNode(4)
var lstNode4 = new ListNode(5)
l1.next = lstNode1
lstNode1.next = lstNode2
lstNode2.next = lstNode3
lstNode3.next = lstNode4

console.log(removeNthFromEnd(l1, 2))

19. 删除链表的倒数第 N 个结点 #

19. 删除链表的倒数第 N 个结点