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。

我在网上找到了一个有趣的解决办法。

礼貌:https://improveandrepeat.com/2018/08/a-simple-way-to-shuffle-your-lists-in-c/

var shuffled = myList。OrderBy(x => Guid.NewGuid()).ToList();

解决这类问题的一个非常简单的方法是在列表中使用一些随机的元素交换。

在伪代码中是这样的:

do 
    r1 = randomPositionInList()
    r2 = randomPositionInList()
    swap elements at index r1 and index r2 
for a certain number of times

    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.

只是想建议使用IComparer<T>和List.Sort()的变体:

public class RandomIntComparer : IComparer<int>
{
    private readonly Random _random = new Random();
    
    public int Compare(int x, int y)
    {
        return _random.Next(-1, 2);
    }
}

用法:

list.Sort(new RandomIntComparer());

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);
}