19. 删除链表的倒数第N个节点

给定一个链表: 1->2->3->4->5, 和 n = 2.

当删除了倒数第二个节点后，链表变为 1->2->3->5.

/*
 * @lc app=leetcode id=19 lang=javascript
 *
 * [19] Remove Nth Node From End of List
 *
 * https://leetcode.com/problems/remove-nth-node-from-end-of-list/description/
 *
 * algorithms
 * Medium (34.03%)
 * Total Accepted:    360.1K
 * Total Submissions: 1.1M
 * Testcase Example:  '[1,2,3,4,5]\n2'
 *
 * Given a linked list, remove the n-th node from the end of list and return
 * its head.
 *
 * Example:
 *
 *
 * Given linked list: 1->2->3->4->5, and n = 2.
 *
 * After removing the second node from the end, the linked list becomes
 * 1->2->3->5.
 *
 *
 * Note:
 *
 * Given n will always be valid.
 *
 * Follow up:
 *
 * Could you do this in one pass?
 *
 */
/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} n
 * @return {ListNode}
 */
var removeNthFromEnd = function(head, n) {
  let i = -1;
  const noop = {
    next: null
  };

  const dummyHead = new ListNode(); // 增加一个dummyHead 简化操作
  dummyHead.next = head;

  let currentP1 = dummyHead;
  let currentP2 = dummyHead;


  while (currentP1) {

    if (i === n) {
      currentP2 = currentP2.next;
    }

    if (i !== n) {
        i++;
    }

    currentP1 = currentP1.next;
  }

  currentP2.next = ((currentP2 || noop).next || noop).next;

  return dummyHead.next;
};