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。

    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.

实现:

public static class ListExtensions
{
    public static void Shuffle<T>(this IList<T> list, Random random)
    {
        for (var i = list.Count - 1; i > 0; i--)
        {
            int indexToSwap = random.Next(i + 1);
            (list[indexToSwap], list[i]) = (list[i], list[indexToSwap]);
        }
    }
}

例子:

var random = new Random();
var array = new [] { 1, 2, 3 };
array.Shuffle(random);
foreach (var item in array) {
    Console.WriteLine(item);
}

.NET小提琴演示

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。

这是我最喜欢的shuffle方法，当不需要修改原始的时候。它是Fisher-Yates“由内到外”算法的变体，适用于任何可枚举序列(源的长度不需要从一开始就知道)。

public static IList<T> NextList<T>(this Random r, IEnumerable<T> source)
{
  var list = new List<T>();
  foreach (var item in source)
  {
    var i = r.Next(list.Count + 1);
    if (i == list.Count)
    {
      list.Add(item);
    }
    else
    {
      var temp = list[i];
      list[i] = item;
      list.Add(temp);
    }
  }
  return list;
}

该算法还可以通过分配一个从0到length - 1的范围来实现，并通过将随机选择的索引与最后一个索引交换来随机耗尽索引，直到所有索引都被选中一次。上面的代码完成了完全相同的事情，但没有额外的分配。非常简洁。

With regards to the Random class it's a general purpose number generator (and If I was running a lottery I'd consider using something different). It also relies on a time based seed value by default. A small alleviation of the problem is to seed the Random class with the RNGCryptoServiceProvider or you could use the RNGCryptoServiceProvider in a method similar to this (see below) to generate uniformly chosen random double floating point values but running a lottery pretty much requires understanding randomness and the nature of the randomness source.

var bytes = new byte[8];
_secureRng.GetBytes(bytes);
var v = BitConverter.ToUInt64(bytes, 0);
return (double)v / ((double)ulong.MaxValue + 1);

生成随机双精度(仅在0和1之间)的目的是用于扩展到整数解。如果你需要从一个基于随机双x的列表中选择一个东西，它总是0 <= x && x < 1是很简单的。

return list[(int)(x * list.Count)];

享受吧!

我通常使用:

var list = new List<T> ();
fillList (list);
var randomizedList = new List<T> ();
var rnd = new Random ();
while (list.Count != 0)
{
    var index = rnd.Next (0, list.Count);
    randomizedList.Add (list [index]);
    list.RemoveAt (index);
}