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);

    public static List<T> Randomize<T>(List<T> list)
    {
        List<T> randomizedList = new List<T>();
        Random rnd = new Random();
        while (list.Count > 0)
        {
            int index = rnd.Next(0, list.Count); //pick a random item from the master list
            randomizedList.Add(list[index]); //place it at the end of the randomized list
            list.RemoveAt(index);
        }
        return randomizedList;
    }

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);

I'm bit surprised by all the clunky versions of this simple algorithm here. Fisher-Yates (or Knuth shuffle) is bit tricky but very compact. Why is it tricky? Because your need to pay attention to whether your random number generator r(a,b) returns value where b is inclusive or exclusive. I've also edited Wikipedia description so people don't blindly follow pseudocode there and create hard to detect bugs. For .Net, Random.Next(a,b) returns number exclusive of b so without further ado, here's how it can be implemented in C#/.Net:

public static void Shuffle<T>(this IList<T> list, Random rnd)
{
    for(var i=list.Count; i > 0; i--)
        list.Swap(0, rnd.Next(0, i));
}

public static void Swap<T>(this IList<T> list, int i, int j)
{
    var temp = list[i];
    list[i] = list[j];
    list[j] = temp;
}

试试这段代码。

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。

如果您有一个固定的数字(75)，您可以创建一个包含75个元素的数组，然后枚举您的列表，将元素移动到数组中的随机位置。您可以使用Fisher-Yates shuffle生成列表号到数组索引的映射。