添加新节点时，链表的浅拷贝不反映更改

I have done a lot of readings but seems like I can't clear my confusion without asking here. Based on the diagram, when I create a a shallow copy of a linkedlist using clone(). A new linkedlist is created and the reference value of the head variable in the original is copied to the clone's and the rest of the nodes are shared. So if I add a new node using the clone, this should be visible to the orginal is it not? But when printing list1 out the value 3 is omitted. Can someone tell me why?

LinkedList<Integer> list1 = new LinkedList<>();
l1.add(1);
l1.add(2);
LinkedList<Integer> list2 = (LinkedList) l1.clone();
l2.add(3); 

clone()创造新的LinkedList结构并返回对第一个节点的新引用。这两者之间的关系LinkedLists 是它们共享同一个节点values。当你做一些add\ remove对旧列表或新列表的操作这些操作不会更改其他列表。这就是我们这样做的原因copy- 当我们更改副本时，我们不想更改原始链表结构。

From LinkedList.clone文档：

返回此内容的浅拷贝LinkedList. (The 元素本身 没有被克隆.) @return这个的浅拷贝LinkedList实例

考虑下面的例子：

import java.util.LinkedList;
import java.util.concurrent.atomic.AtomicInteger;

public class LinkedListsApp {

    public static void main(String[] args) throws Exception {
        LinkedList<AtomicInteger> l1 = new LinkedList<>();
        l1.add(new AtomicInteger(100));
        l1.add(new AtomicInteger(200));

        LinkedList<AtomicInteger> l2 = (LinkedList) l1.clone();
        l2.add(new AtomicInteger(300));

        System.out.println(l1);
        System.out.println(l2);

        // change element on first list
        l1.get(0).incrementAndGet();

        System.out.println();
        System.out.println("After change internal state of first element");
        System.out.println(l1);
        System.out.println(l2);
    }
}

上面的代码打印：

[100, 200]
[100, 200, 300]

After change internal state of first element
[101, 200]
[101, 200, 300]

正如我们所看到的，当我们更改第一个列表中第一个元素的内部状态时，它对于第二个列表也是可见的。因此，不存在每个元素值的深层复制，而是结构的复制 - 节点和顺序的复制。

为了完全清楚，让我们看一下实施Java 8:

public Object clone() {
    LinkedList<E> clone = superClone();

    // Put clone into "virgin" state
    clone.first = clone.last = null;
    clone.size = 0;
    clone.modCount = 0;

    // Initialize clone with our elements
    for (Node<E> x = first; x != null; x = x.next)
        clone.add(x.item);

    return clone;
}

看看for-each环形。它迭代原始列表并将值添加到clone列表。方法add创造新的Node存储与原始列表相同值的对象：x.item.