我同意上面提到的大部分观点。我也同意，在大多数情况下，List看起来是一个更明显的选择。

但是，我只是想补充一点，在很多情况下，LinkedList比List更有效。

假设你正在遍历元素，你想要执行大量的插入/删除;LinkedList在线性O(n)时间内完成，而List在二次O(n²)时间内完成。 假设你想一次又一次地访问更大的对象，LinkedList就变得非常有用。 Deque()和queue()更好地使用LinkedList实现。 当你处理很多更大的对象时，增加LinkedList的大小会更容易和更好。

希望有人会觉得这些评论有用。

我之前的回答不够准确。 D是正确答案 但现在我可以发布更有用和正确的答案。

我做了一些额外的检查您可以通过以下链接找到它的源代码，并在您自己的环境中通过https://github.com/ukushu/DataStructuresTestsAndOther.git重新检查它

短的结果:

Array need to use: So often as possible. It's fast and takes smallest RAM range for same amount information. If you know exact count of cells needed If data saved in array < 85000 b (85000/32 = 2656 elements for integer data) If needed high Random Access speed List need to use: If needed to add cells to the end of list (often) If needed to add cells in the beginning/middle of the list (NOT OFTEN) If data saved in array < 85000 b (85000/32 = 2656 elements for integer data) If needed high Random Access speed LinkedList need to use: If needed to add cells in the beginning/middle/end of the list (often) If needed only sequential access (forward/backward) If you need to save LARGE items, but items count is low. Better do not use for large amount of items, as it's use additional memory for links.

更多的细节:

有趣的是:

LinkedList<T> internally is not a List in .NET. It's even does not implement IList<T>. And that's why there are absent indexes and methods related to indexes. LinkedList<T> is node-pointer based collection. In .NET it's in doubly linked implementation. This means that prior/next elements have link to current element. And data is fragmented -- different list objects can be located in different places of RAM. Also there will be more memory used for LinkedList<T> than for List<T> or Array. List<T> in .Net is Java's alternative of ArrayList<T>. This means that this is array wrapper. So it's allocated in memory as one contiguous block of data. If allocated data size exceeds 85000 bytes, it will be moved to Large Object Heap. Depending on the size, this can lead to heap fragmentation(a mild form of memory leak). But in the same time if size < 85000 bytes -- this provides a very compact and fast-access representation in memory. Single contiguous block is preferred for random access performance and memory consumption but for collections that need to change size regularly a structure such as an Array generally need to be copied to a new location whereas a linked list only needs to manage the memory for the newly inserted/deleted nodes.

List和LinkedList之间的区别在于它们的底层实现。List是基于数组的集合(ArrayList)。LinkedList是基于节点指针的集合(LinkedListNode)。在API级别的使用上，它们几乎是相同的，因为它们都实现了相同的接口集，如ICollection、IEnumerable等。

关键的区别在于性能问题。例如，如果您正在实现具有大量“INSERT”操作的列表，LinkedList的性能优于list。因为LinkedList可以在O(1)时间内完成，但是List可能需要扩展底层数组的大小。要了解更多信息/细节，你可能想要阅读LinkedList和数组数据结构之间的算法差异。http://en.wikipedia.org/wiki/Linked_list和Array

希望这能有所帮助，

将链表视为列表可能会有一些误导。它更像一个链条。事实上，在。net中，LinkedList<T>甚至没有实现IList<T>。在链表中没有真正的索引概念，即使看起来有。当然，类中提供的方法都不接受索引。

链表可以是单链，也可以是双链。这是指链中的每个元素是否只与下一个元素有链接(单链)，还是与前一个/下一个元素都有链接(双链)。LinkedList<T>是双重链接。

在内部，List<T>由一个数组支持。这在内存中提供了一个非常紧凑的表示。相反，LinkedList<T>涉及额外的内存来存储连续元素之间的双向链接。因此LinkedList<T>的内存占用通常会比List<T>的内存占用大(需要注意的是，List<T>可以有未使用的内部数组元素，以提高追加操作期间的性能)。

它们也有不同的表现特征:

附加

LinkedList<T>.AddLast(item)常量时间 List<T>.Add(item)平摊常数时间，线性最坏情况

预谋

LinkedList<T>.AddFirst(item)常量时间 列表> < T。插入(0，项)线性时间

插入

LinkedList < T >。AddBefore(节点，项目)常量时间 LinkedList < T >。AddAfter(节点，项目)常量时间 列表> < T。插入(索引、项)线性时间

删除

删除(项目)线性时间 删除(节点)常量时间 列表<T>.删除(项目)线性时间 List<T>.RemoveAt(index)线性时间

数

LinkedList < T >。计算常数时间 列表> < T。计算常数时间

包含

包含(项目)线性时间 列表<T>.包含(项)线性时间

清晰的

LinkedList<T>.Clear()线性时间 List<T>.Clear()线性时间

如你所见，它们基本上是相等的。实际上，LinkedList<T>的API使用起来更麻烦，其内部需求的细节会泄漏到代码中。

然而，如果你需要在一个列表中做很多插入/删除，它提供了常数时间。List<T>提供线性时间，因为列表中的额外项必须在插入/删除之后重新排列。

这是改编自Tono Nam的公认的答案，纠正了一些错误的测量。

测试:

static void Main()
{
    LinkedListPerformance.AddFirst_List(); // 12028 ms
    LinkedListPerformance.AddFirst_LinkedList(); // 33 ms

    LinkedListPerformance.AddLast_List(); // 33 ms
    LinkedListPerformance.AddLast_LinkedList(); // 32 ms

    LinkedListPerformance.Enumerate_List(); // 1.08 ms
    LinkedListPerformance.Enumerate_LinkedList(); // 3.4 ms

    //I tried below as fun exercise - not very meaningful, see code
    //sort of equivalent to insertion when having the reference to middle node

    LinkedListPerformance.AddMiddle_List(); // 5724 ms
    LinkedListPerformance.AddMiddle_LinkedList1(); // 36 ms
    LinkedListPerformance.AddMiddle_LinkedList2(); // 32 ms
    LinkedListPerformance.AddMiddle_LinkedList3(); // 454 ms

    Environment.Exit(-1);
}

代码是:

using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;

namespace stackoverflow
{
    static class LinkedListPerformance
    {
        class Temp
        {
            public decimal A, B, C, D;

            public Temp(decimal a, decimal b, decimal c, decimal d)
            {
                A = a; B = b; C = c; D = d;
            }
        }



        static readonly int start = 0;
        static readonly int end = 123456;
        static readonly IEnumerable<Temp> query = Enumerable.Range(start, end - start).Select(temp);

        static Temp temp(int i)
        {
            return new Temp(i, i, i, i);
        }

        static void StopAndPrint(this Stopwatch watch)
        {
            watch.Stop();
            Console.WriteLine(watch.Elapsed.TotalMilliseconds);
        }

        public static void AddFirst_List()
        {
            var list = new List<Temp>();
            var watch = Stopwatch.StartNew();

            for (var i = start; i < end; i++)
                list.Insert(0, temp(i));

            watch.StopAndPrint();
        }

        public static void AddFirst_LinkedList()
        {
            var list = new LinkedList<Temp>();
            var watch = Stopwatch.StartNew();

            for (int i = start; i < end; i++)
                list.AddFirst(temp(i));

            watch.StopAndPrint();
        }

        public static void AddLast_List()
        {
            var list = new List<Temp>();
            var watch = Stopwatch.StartNew();

            for (var i = start; i < end; i++)
                list.Add(temp(i));

            watch.StopAndPrint();
        }

        public static void AddLast_LinkedList()
        {
            var list = new LinkedList<Temp>();
            var watch = Stopwatch.StartNew();

            for (int i = start; i < end; i++)
                list.AddLast(temp(i));

            watch.StopAndPrint();
        }

        public static void Enumerate_List()
        {
            var list = new List<Temp>(query);
            var watch = Stopwatch.StartNew();

            foreach (var item in list)
            {

            }

            watch.StopAndPrint();
        }

        public static void Enumerate_LinkedList()
        {
            var list = new LinkedList<Temp>(query);
            var watch = Stopwatch.StartNew();

            foreach (var item in list)
            {

            }

            watch.StopAndPrint();
        }

        //for the fun of it, I tried to time inserting to the middle of 
        //linked list - this is by no means a realistic scenario! or may be 
        //these make sense if you assume you have the reference to middle node

        //insertion to the middle of list
        public static void AddMiddle_List()
        {
            var list = new List<Temp>();
            var watch = Stopwatch.StartNew();

            for (var i = start; i < end; i++)
                list.Insert(list.Count / 2, temp(i));

            watch.StopAndPrint();
        }

        //insertion in linked list in such a fashion that 
        //it has the same effect as inserting into the middle of list
        public static void AddMiddle_LinkedList1()
        {
            var list = new LinkedList<Temp>();
            var watch = Stopwatch.StartNew();

            LinkedListNode<Temp> evenNode = null, oddNode = null;
            for (int i = start; i < end; i++)
            {
                if (list.Count == 0)
                    oddNode = evenNode = list.AddLast(temp(i));
                else
                    if (list.Count % 2 == 1)
                        oddNode = list.AddBefore(evenNode, temp(i));
                    else
                        evenNode = list.AddAfter(oddNode, temp(i));
            }

            watch.StopAndPrint();
        }

        //another hacky way
        public static void AddMiddle_LinkedList2()
        {
            var list = new LinkedList<Temp>();
            var watch = Stopwatch.StartNew();

            for (var i = start + 1; i < end; i += 2)
                list.AddLast(temp(i));
            for (int i = end - 2; i >= 0; i -= 2)
                list.AddLast(temp(i));

            watch.StopAndPrint();
        }

        //OP's original more sensible approach, but I tried to filter out
        //the intermediate iteration cost in finding the middle node.
        public static void AddMiddle_LinkedList3()
        {
            var list = new LinkedList<Temp>();
            var watch = Stopwatch.StartNew();

            for (var i = start; i < end; i++)
            {
                if (list.Count == 0)
                    list.AddLast(temp(i));
                else
                {
                    watch.Stop();
                    var curNode = list.First;
                    for (var j = 0; j < list.Count / 2; j++)
                        curNode = curNode.Next;
                    watch.Start();

                    list.AddBefore(curNode, temp(i));
                }
            }

            watch.StopAndPrint();
        }
    }
}

你可以看到结果与其他人在这里记录的理论性能是一致的。很清楚- LinkedList<T>在插入的情况下获得了很大的时间。我还没有测试从列表中间删除，但结果应该是相同的。当然，List<T>在其他方面表现得更好，比如O(1)随机访问。

在. net中，列表被表示为数组。因此，与LinkedList相比，使用普通List会更快。这就是为什么上面的人看到他们看到的结果。

Why should you use the List? I would say it depends. List creates 4 elements if you don't have any specified. The moment you exceed this limit, it copies stuff to a new array, leaving the old one in the hands of the garbage collector. It then doubles the size. In this case, it creates a new array with 8 elements. Imagine having a list with 1 million elements, and you add 1 more. It will essentially create a whole new array with double the size you need. The new array would be with 2Mil capacity however, you only needed 1Mil and 1. Essentially leaving stuff behind in GEN2 for the garbage collector and so on. So it can actually end up being a huge bottleneck. You should be careful about that.