下面是一个有效的Fisher-Yates shuffle数组函数:

private static void shuffleArray(int[] array)
{
    int index;
    Random random = new Random();
    for (int i = array.length - 1; i > 0; i--)
    {
        index = random.nextInt(i + 1);
        if (index != i)
        {
            array[index] ^= array[i];
            array[i] ^= array[index];
            array[index] ^= array[i];
        }
    }
}

or

private static void shuffleArray(int[] array)
{
    int index, temp;
    Random random = new Random();
    for (int i = array.length - 1; i > 0; i--)
    {
        index = random.nextInt(i + 1);
        temp = array[index];
        array[index] = array[i];
        array[i] = temp;
    }
}

无随机解:

   static void randomArrTimest(int[] some){
        long startTime = System.currentTimeMillis();
        for (int i = 0; i < some.length; i++) {
            long indexToSwap = startTime%(i+1);
            long tmp = some[(int) indexToSwap];
            some[(int) indexToSwap] = some[i];
            some[i] = (int) tmp;
        }
        System.out.println(Arrays.toString(some));
    }

下面是一个使用数组列表的简单方法:

List<Integer> solution = new ArrayList<>();
for (int i = 1; i <= 6; i++) {
    solution.add(i);
}
Collections.shuffle(solution);

使用ArrayList<Integer>可以帮助您解决洗牌问题，而无需应用大量逻辑和消耗更少的时间。以下是我的建议:

ArrayList<Integer> x = new ArrayList<Integer>();
for(int i=1; i<=add.length(); i++)
{
    x.add(i);
}
Collections.shuffle(x);

你应该使用Collections.shuffle()。但是，不能直接操作原始类型数组，因此需要创建包装器类。

试试这个。

public static void shuffle(int[] array) {
    Collections.shuffle(new AbstractList<Integer>() {
        @Override public Integer get(int index) { return array[index]; }
        @Override public int size() { return array.length; }
        @Override public Integer set(int index, Integer element) {
            int result = array[index];
            array[index] = element;
            return result;
        }
    });
}

And

int[] solutionArray = {1, 2, 3, 4, 5, 6, 6, 5, 4, 3, 2, 1};
shuffle(solutionArray);
System.out.println(Arrays.toString(solutionArray));

输出:

[3, 3, 4, 1, 6, 2, 2, 1, 5, 6, 5, 4]

类似的情况没有使用swap b

    Random r = new Random();
    int n = solutionArray.length;
    List<Integer> arr =  Arrays.stream(solutionArray)
                               .boxed()
                               .collect(Collectors.toList());
    for (int i = 0; i < n-1; i++) {
        solutionArray[i] = arr.remove(r.nextInt(arr.size())); // randomize based on size
    }
    solutionArray[n-1] = arr.get(0);