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));
    }

下面是数组的泛型版本:

import java.util.Random;

public class Shuffle<T> {

    private final Random rnd;

    public Shuffle() {
        rnd = new Random();
    }

    /**
     * Fisher–Yates shuffle.
     */
    public void shuffle(T[] ar) {
        for (int i = ar.length - 1; i > 0; i--) {
            int index = rnd.nextInt(i + 1);
            T a = ar[index];
            ar[index] = ar[i];
            ar[i] = a;
        }
    }
}

考虑到ArrayList基本上只是一个数组，使用ArrayList而不是显式数组并使用Collections.shuffle()可能是明智的。但是，性能测试并没有显示上述方法与Collections.sort()之间有任何显著差异:

Shuffe<Integer>.shuffle(...) performance: 576084 shuffles per second
Collections.shuffle(ArrayList<Integer>) performance: 629400 shuffles per second
MathArrays.shuffle(int[]) performance: 53062 shuffles per second

Apache Commons实现MathArrays。Shuffle限制为int[]，性能损失可能是由于使用了随机数生成器。

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]

还有另一种方法，还没有发布

//that way, send many object types diferentes
public anotherWayToReciveParameter(Object... objects)
{
    //ready with array
    final int length =objects.length;
    System.out.println(length);
    //for ready same list
    Arrays.asList(objects);
}

这种方法更简单，取决于上下文

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);

使用集合来洗牌一个基本类型数组有点过分……

你自己实现这个函数很简单，比如使用Fisher-Yates shuffle:

import java.util.*;
import java.util.concurrent.ThreadLocalRandom;

class Test
{
  public static void main(String args[])
  {
    int[] solutionArray = { 1, 2, 3, 4, 5, 6, 16, 15, 14, 13, 12, 11 };

    shuffleArray(solutionArray);
    for (int i = 0; i < solutionArray.length; i++)
    {
      System.out.print(solutionArray[i] + " ");
    }
    System.out.println();
  }

  // Implementing Fisher–Yates shuffle
  static void shuffleArray(int[] ar)
  {
    // If running on Java 6 or older, use `new Random()` on RHS here
    Random rnd = ThreadLocalRandom.current();
    for (int i = ar.length - 1; i > 0; i--)
    {
      int index = rnd.nextInt(i + 1);
      // Simple swap
      int a = ar[index];
      ar[index] = ar[i];
      ar[i] = a;
    }
  }
}