现在聊LinkedList的peek操作，这个操作都是进行输出参数，不会删除相关Node的。

/**
 * Retrieves, but does not remove, the first element of this list,
 * or returns {@code null} if this list is empty.
 *
 * @return the first element of this list, or {@code null}
 *         if this list is empty
 * @since 1.6
 */
public E peekFirst() {
    final Node<E> f = first;
    return (f == null) ? null : f.item;
 }
/**
 * Retrieves, but does not remove, the last element of this list,
 * or returns {@code null} if this list is empty.
 *
 * @return the last element of this list, or {@code null}
 *         if this list is empty
 * @since 1.6
 */
public E peekLast() {
    final Node<E> l = last;
    return (l == null) ? null : l.item;
}

peek()的代码和peekFirst的代码是相同的都是取第一个数据。

然后是poll()操作，这个是将数据拿出然后删除该Node节点的。

/**
     * Retrieves and removes the head (first element) of this list.
     *
     * @return the head of this list, or {@code null} if this list is empty
     * @since 1.5
     */
    public E poll() {
        final Node<E> f = first;
        return (f == null) ? null : unlinkFirst(f);
    }

程序回跳转到unlikFirst()的方法：

/**
     * Unlinks non-null first node f.
     */
    private E unlinkFirst(Node<E> f) {
        // assert f == first && f != null;
        final E element = f.item;
        final Node<E> next = f.next;
        f.item = null;
        f.next = null; // help GC
        first = next;
        if (next == null)
            last = null;
        else
            next.prev = null;
        size--;
        modCount++;
        return element;
    }

poll和pollFirst的代码相同都是除去第一个，pollLast是除去最后一个，跳转到unlikLast的代码其内容和unlikFirst基本相同流程。

list.pop(),是直接移除改Node节点 并没有返回该节点的数据。

public E pop() {
    return removeFirst();
}

push(e)方法就是向数据中添加一个新的Node节点：

public void push(E e) {
    addFirst(e);
}

get方法相关的有get(index),getFirst(),getLast(),

/**
 * Returns the element at the specified position in this list.
 *
 * @param index index of the element to return
 * @return the element at the specified position in this list
 * @throws IndexOutOfBoundsException {@inheritDoc}
 */
public E get(int index) {
    checkElementIndex(index);
    return node(index).item;
}
private void checkElementIndex(int index) {
    if (!isElementIndex(index))
        throw new IndexOutOfBoundsException(outOfBoundsMsg(index));
}
/**
 * Tells if the argument is the index of an existing element.
 */
private boolean isElementIndex(int index) {
    return index >= 0 && index < size;
}

获取第index位置的Node节点数据时，先进行对该节点位置进行判断是否有效，然后进行获取该位置数据：

/**
 * Returns the (non-null) Node at the specified element index.
 */
Node<E> node(int index) {
    // assert isElementIndex(index);

    if (index < (size >> 1)) {
        Node<E> x = first;
        for (int i = 0; i < index; i++)
            x = x.next;
        return x;
    } else {
        Node<E> x = last;
        for (int i = size - 1; i > index; i--)
            x = x.prev;
        return x;
    }
}

获取第一个或者最后一个的数据时：

/**
 * Returns the first element in this list.
 *
 * @return the first element in this list
 * @throws NoSuchElementException if this list is empty
 */
public E getFirst() {
    final Node<E> f = first;
    if (f == null)
        throw new NoSuchElementException();
    return f.item;
}

Node直接指向first或者last从而直接获取item。 list.element()也是获取第一个Node的数据：

/**
 * Retrieves, but does not remove, the head (first element) of this list.
 *
 * @return the head of this list
 * @throws NoSuchElementException if this list is empty
 * @since 1.5
 */
public E element() {
    return getFirst();
}