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。

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。

当您需要内置索引访问、排序(以及二进制搜索之后)和“ToArray()”方法时，您应该使用List。

在大多数情况下，List<T>更有用。LinkedList<T>在列表中间添加/删除项时成本更低，而list <T>只能在列表末尾添加/删除项。

LinkedList<T>只有在访问顺序数据(向前或向后)时才最有效-随机访问相对昂贵，因为它每次都必须遍历链(因此它没有索引器)。但是，因为List<T>本质上只是一个数组(带有包装器)，所以随机访问是可以的。

List<T>还提供了很多支持方法- Find, ToArray等;然而，这些也可以通过扩展方法用于。net 3.5/ c# 3.0的LinkedList<T> -所以这不是一个因素。

使用LinkedList的常见情况是这样的:

假设您想要从一个字符串列表中删除许多特定的字符串，这些字符串的大小很大，比如100,000。要删除的字符串可以在HashSet dic中查找，字符串列表中应该包含30,000到60,000个这样的需要删除的字符串。

那么用于存储100,000个字符串的列表的最佳类型是什么?答案是LinkedList。如果它们存储在数组列表中，则遍历它并删除匹配的字符串将占用 到数十亿次操作，而使用迭代器和remove()方法只需要大约100,000次操作。

LinkedList<String> strings = readStrings();
HashSet<String> dic = readDic();
Iterator<String> iterator = strings.iterator();
while (iterator.hasNext()){
    String string = iterator.next();
    if (dic.contains(string))
    iterator.remove();
}

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.