链表_19删除链表的倒数第N个节点.md

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。

进阶：你能尝试使用一趟扫描实现吗？

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]

法一： 快慢指针
var removeNthFromEnd = function (head, n) {
  let dummyHead = new ListNode(0, head);
  let slow = dummyHead;
  let fast = dummyHead;
  while (n-- && fast) {
    fast = fast.next
  }
  fast = fast.next; // fast再提前走一步，因为需要让slow指向删除节点的上一个节点
  while (fast) {
    fast = fast.next;
    slow = slow.next;
  }
  slow.next = slow.next.next

  return dummyHead.next;
};

法二： 先遍历得到正序是第几个元素移除，再次遍历移除太严肃
var removeNthFromEnd = function (head, n) {
  let dummyHead = new ListNode(0, head);
  let cur1 = dummyHead;
  let cur = dummyHead;

  let num1 = 0;
  let num = 0;

  while (cur1) {
    cur1 = cur1.next
    num1++;
  }

  while (cur) {
    if (num === num1 - n -1) {
      cur.next = cur.next.next
      console.log(num)
    } else {
      cur = cur.next
    }
    num++;
  }
  console.log(cur1, cur, dummyHead)

  return dummyHead.next;
};