List<T> OriginalList = new List<T>();
    List<T> TempList = new List<T>();
    Random random = new Random();
    int length = OriginalList.Count;
    int TempIndex = 0;

    while (length > 0) {
        TempIndex = random.Next(0, length);  // get random value between 0 and original length
        TempList.Add(OriginalList[TempIndex]); // add to temp list
        OriginalList.RemoveAt(TempIndex); // remove from original list
        length = OriginalList.Count;  // get new list <T> length.
    }

    OriginalList = new List<T>();
    OriginalList = TempList; // copy all items from temp list to original list.

如果你不介意使用两个list，那么这可能是最简单的方法，但可能不是最有效或最不可预测的方法:

List<int> xList = new List<int>() { 1, 2, 3, 4, 5 };
List<int> deck = new List<int>();

foreach (int xInt in xList)
    deck.Insert(random.Next(0, deck.Count + 1), xInt);

我们可以为List使用扩展方法，并使用线程安全的随机生成器组合。我在NuGet上打包了一个改进的版本，并在GitHub上提供源代码。NuGet版本包含可选的密码强随机。

Pre -。NET 6.0版本:

[MethodImpl(MethodImplOptions.AggressiveInlining)]
public static void Shuffle<T>(this IList<T> list)
{
    if (list == null) throw new ArgumentNullException(nameof(list));
    int n = list.Count;
    while (n > 1)
    {
        int k = ThreadSafeRandom.Instance.Next(n--);
        (list[n], list[k]) = (list[k], list[n]);
    }
}

internal class ThreadSafeRandom
{
    public static Random Instance => _local.Value;

    private static readonly Random _global = new Random();
    private static readonly ThreadLocal<Random> _local = new ThreadLocal<Random>(() =>
    {
        int seed;
        lock (_global)
        {
            seed = _global.Next();
        }
        return new Random(seed);
    });
}

在。net 6.0或更高版本上:

[MethodImpl(MethodImplOptions.AggressiveInlining)]
public static void Shuffle<T>(this IList<T> list)
{
    ArgumentNullException.ThrowIfNull(list);
    int n = list.Count;
    while (n > 1)
    {
        int k = Random.Shared.Next(n--);
        (list[n], list[k]) = (list[k], list[n]);
    }
}

通过NuGet安装库以获得更多功能。

如果我们只需要以完全随机的顺序洗牌项目(只是在一个列表中混合项目)，我更喜欢这个简单而有效的代码，按guid排序项目…

var shuffledcards = cards.OrderBy(a => Guid.NewGuid()).ToList();

正如人们在评论中指出的那样，guid不能保证是随机的，所以我们应该使用真正的随机数生成器:

private static Random rng = new Random();
...
var shuffledcards = cards.OrderBy(a => rng.Next()).ToList();

对已接受答案的简单修改，返回一个新的列表，而不是原地工作，并像许多其他Linq方法一样接受更通用的IEnumerable<T>。

private static Random rng = new Random();

/// <summary>
/// Returns a new list where the elements are randomly shuffled.
/// Based on the Fisher-Yates shuffle, which has O(n) complexity.
/// </summary>
public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> list) {
    var source = list.ToList();
    int n = source.Count;
    var shuffled = new List<T>(n);
    shuffled.AddRange(source);
    while (n > 1) {
        n--;
        int k = rng.Next(n + 1);
        T value = shuffled[k];
        shuffled[k] = shuffled[n];
        shuffled[n] = value;
    }
    return shuffled;
}

下面是一个高效的Shuffler，它返回一个字节数组的打乱值。它从来不会超过需要的次数。它可以从之前停止的地方重新启动。我的实际实现(未显示)是一个MEF组件，它允许用户指定替换洗牌器。

    public byte[] Shuffle(byte[] array, int start, int count)
    {
        int n = array.Length - start;
        byte[] shuffled = new byte[count];
        for(int i = 0; i < count; i++, start++)
        {
            int k = UniformRandomGenerator.Next(n--) + start;
            shuffled[i] = array[k];
            array[k] = array[start];
            array[start] = shuffled[i];
        }
        return shuffled;
    }

`