/**
  * Appends the specified element to the end of this list.
  *
  * 
This method is equivalent to {@link #addLast}.
  *
  * @param e element to be appended to this list
  * @return {@code true} (as specified by {@link Collection#add})
  */
 public boolean add(E e) {
  linkLast(e);
  return true;
 }
 
 /**
  * Links e as last element.
  */
 void linkLast(E e) {
  final Node l = last;        // 将当前最后一个元素寄存在 l
  final Node newNode = new Node<>(l, e, null);  // new 一个新节点：pre的引用为l；存储元素为e；next的引用为null
  last = newNode;          // 将新节点引用覆盖成员变量 last
  if (l == null)          
   first = newNode;        // 若l为null，说明之前链表为空，此时新节点为首个元素
  else
   l.next = newNode;        // 否则，更新l的next引用
  size++;            // size+1
  modCount++;           // 非查询操作 modCount 都会 +1
 }

/**
  * Inserts the specified element at the specified position in this list.
  * Shifts the element currently at that position (if any) and any
  * subsequent elements to the right (adds one to their indices).
  *
  * @param index index at which the specified element is to be inserted
  * @param element element to be inserted
  * @throws IndexOutOfBoundsException {@inheritDoc}
  */
 public void add(int index, E element) {
  checkPositionIndex(index); // 检查 index 是否大于 size

  if (index == size)
   linkLast(element);  // 直接在链表末尾追加
  else
   linkBefore(element, node(index)); // 插入index 节点前面
 }
 
 
 // 检查 index 是否超出范围 超出则抛出 IndexOutOfBoundsException
 private void checkPositionIndex(int index) {
  if (!isPositionIndex(index))
   throw new IndexOutOfBoundsException(outOfBoundsMsg(index));
 }

 /**
  * Tells if the argument is the index of a valid position for an
  * iterator or an add operation.
  */
 private boolean isPositionIndex(int index) {
  return index >= 0 && index <= size;
 }
 
 
 
 /**
  * 根据 index 查找 node
  * 该方法利用了双向链表的特性，index 距离哪个链表头近就从哪边开始开始遍历
  * 时间复杂度为 O(n/2)；
  * 当 index 接近 size 的中间值时，效率最低
  * Returns the (non-null) Node at the specified element index.
  */
 Node node(int index) {
  // assert isElementIndex(index);

  if (index < (size >> 1)) {   // size 右移一位（除以2）
   Node x = first;
   for (int i = 0; i < index; i++)
    x = x.next;
   return x;
  } else {
   Node x = last;
   for (int i = size - 1; i > index; i--)
    x = x.prev;
   return x;
  }
 }