假设您有一个有序集，您还想通过添加和删除元素来修改它。此外，您需要能够以这样的方式保留对元素的引用，以便稍后可以获得前一个或下一个元素。例如，一本书中的待办事项列表或一组段落。

首先，我们应该注意到，如果您想在集合本身之外保留对对象的引用，那么您可能最终会将指针存储在数组中，而不是存储对象本身。否则你将无法插入到数组中-如果对象嵌入到数组中，它们将在插入期间移动，并且任何指向它们的指针都将无效。数组下标也是如此。

您的第一个问题，正如您自己所注意到的，是插入链表允许插入O(1)，但数组通常需要O(n)。这个问题可以部分克服——可以创建一种数据结构，提供类似数组的按顺序访问接口，其中读和写在最坏的情况下都是对数的。

Your second, and more severe problem is that given an element finding next element is O(n). If the set was not modified you could retain the index of the element as the reference instead of the pointer thus making find-next an O(1) operation, but as it is all you have is a pointer to the object itself and no way to determine its current index in the array other than by scanning the entire "array". This is an insurmountable problem for arrays - even if you can optimized insertions, there is nothing you can do to optimize find-next type operation.

Eric Lippert最近发表了一篇关于数组应该保守使用的原因之一的文章。

使用链接列表的人必须阅读。人们会再次爱上数组的。 它谈到了 无序执行，硬件预取，内存延迟等。

http://www.futurechips.org/thoughts-for-researchers/quick-post-linked-lists.html

It's easier to store data of different sizes in a linked list. An array assumes every element is exactly the same size. As you mentioned, it's easier for a linked list to grow organically. An array's size needs to be known ahead of time, or re-created when it needs to grow. Shuffling a linked list is just a matter of changing what points to what. Shuffling an array is more complicated and/or takes more memory. As long as your iterations all happen in a "foreach" context, you don't lose any performance in iteration.

维基百科上有很好的章节介绍了它们的区别。

Linked lists have several advantages over arrays. Elements can be inserted into linked lists indefinitely, while an array will eventually either fill up or need to be resized, an expensive operation that may not even be possible if memory is fragmented. Similarly, an array from which many elements are removed may become wastefully empty or need to be made smaller. On the other hand, arrays allow random access, while linked lists allow only sequential access to elements. Singly-linked lists, in fact, can only be traversed in one direction. This makes linked lists unsuitable for applications where it's useful to look up an element by its index quickly, such as heapsort. Sequential access on arrays is also faster than on linked lists on many machines due to locality of reference and data caches. Linked lists receive almost no benefit from the cache. Another disadvantage of linked lists is the extra storage needed for references, which often makes them impractical for lists of small data items such as characters or boolean values. It can also be slow, and with a naïve allocator, wasteful, to allocate memory separately for each new element, a problem generally solved using memory pools.

http://en.wikipedia.org/wiki/Linked_list

A widely unappreciated argument for ArrayList and against LinkedList is that LinkedLists are uncomfortable while debugging. The time spent by maintenance developers to understand the program, e.g. to find bugs, increases and IMHO does sometimes not justify the nanoseconds in performance improvements or bytes in memory consumption in enterprise applicatons. Sometimes (well, of course it depends on the type of applications), it's better to waste a few bytes but have an application which is more maintainable or easier to understand.

例如，在Java环境中，使用Eclipse调试器，调试ArrayList将显示一个非常容易理解的结构:

arrayList   ArrayList<String>
  elementData   Object[]
    [0] Object  "Foo"
    [1] Object  "Foo"
    [2] Object  "Foo"
    [3] Object  "Foo"
    [4] Object  "Foo"
    ...

另一方面，查看LinkedList的内容并找到特定的对象变成了一个展开树的噩梦，更不用说过滤LinkedList内部信息所需的认知开销:

linkedList  LinkedList<String>
    header  LinkedList$Entry<E>
        element E
        next    LinkedList$Entry<E>
            element E   "Foo"
            next    LinkedList$Entry<E>
                element E   "Foo"
                next    LinkedList$Entry<E>
                    element E   "Foo"
                    next    LinkedList$Entry<E>
                    previous    LinkedList$Entry<E>
                    ...
                previous    LinkedList$Entry<E>
            previous    LinkedList$Entry<E>
        previous    LinkedList$Entry<E>