需要使用密钥进行可重复的随机数组随机播放

问题描述

我正在寻找使用键随机地整理列表/数组的方法。我希望能够使用该键重复相同的随机顺序。

I am looking to randomly shuffle a list/array using a key. I want to be able to repeat the same random order using the key.

所以我将随机生成一个数字键，例如1到20，然后使用该键尝试

So I will randomly generate a numeric key from say 1 to 20 then use that key to try and randomly shuffle the list.

我首先尝试使用键来不断遍历我的列表，将键递减到0，然后抓住我所在的任何元素，将其删除并将其添加到我的随机数组中。结果是随机的，但是当数组很小（我的大部分将是）和/或键很小时，它最终并不会改组……似乎更多的是移位。

I first tried just using the key to keep iterating through my list, decrementing the key until=0, then grabbing whatever element I am on, removing it and adding it to my shuffled array. The result is kind of random but when the arrays are small (which most of mine will be) and/or the key is small it doesn't end up shuffling... seems to be more of a shift.

我必须能够确定

这是csharp中的一些示例代码：

Here is some sample code in csharp of :

public static TList<VoteSetupAnswer> ShuffleListWithKey(TList<VoteSetupAnswer> UnsortedList, int ShuffleKey)
    {
        TList<VoteSetupAnswer> SortedList = new TList<VoteSetupAnswer>();
        int UnsortedListCount = UnsortedList.Count;
        for (int i = 0; i < UnsortedListCount; i++)
        {
            int Location;
            SortedList.Add(OneArrayCycle(UnsortedList, ShuffleKey, out Location));
            UnsortedList.RemoveAt(Location);
        }
        return SortedList;
    }

    public static VoteSetupAnswer OneArrayCycle(TList<VoteSetupAnswer> array, int ShuffleKey, out int Location)
    {
        Location = 0;
        if (ShuffleKey == 1)
        {
            Location = 0;
            return array[0];
        }
        else
        {
            for (int x = 0; x <= ShuffleKey; x++)
            {
                if (x == ShuffleKey)
                    return array[Location];
                Location++;
                if (Location == array.Count)
                    Location = 0;
            }
            return array[Location];
        }
    }

推荐答案

 /**
     * Randomly permutes the array of this permutation. All permutations occur with approximately equal
     * likelihood. This implementation traverses the permutation array forward, from the first element up to
     * the second last, repeatedly swapping a randomly selected element into the "current position". Elements
     * are randomly selected from the portion of the permutation array that runs from the current position to
     * the last element, inclusive.
     * <p>
     * This method runs in linear time.
     */
    public static void shuffle(Random random, int[] a) {
        for (int i = 0; i < a.length - 1; i++) {
            swap(a, i, i + random.nextInt(a.length - i));
        }
    }

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