我在权衡这个非常流行的问题，因为没有人写过一个shuffle-copy版本。样式大量借鉴了Arrays.java，因为现在谁没有掠夺Java技术呢?包括泛型和int实现。

   /**
    * Shuffles elements from {@code original} into a newly created array.
    *
    * @param original the original array
    * @return the new, shuffled array
    * @throws NullPointerException if {@code original == null}
    */
   @SuppressWarnings("unchecked")
   public static <T> T[] shuffledCopy(T[] original) {
      int originalLength = original.length; // For exception priority compatibility.
      Random random = new Random();
      T[] result = (T[]) Array.newInstance(original.getClass().getComponentType(), originalLength);

      for (int i = 0; i < originalLength; i++) {
         int j = random.nextInt(i+1);
         result[i] = result[j];
         result[j] = original[i];
      }

      return result;
   }


   /**
    * Shuffles elements from {@code original} into a newly created array.
    *
    * @param original the original array
    * @return the new, shuffled array
    * @throws NullPointerException if {@code original == null}
    */
   public static int[] shuffledCopy(int[] original) {
      int originalLength = original.length;
      Random random = new Random();
      int[] result = new int[originalLength];

      for (int i = 0; i < originalLength; i++) {
         int j = random.nextInt(i+1);
         result[i] = result[j];
         result[j] = original[i];
      }

      return result;
   }

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。

Random rnd = new Random();
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;
}

顺便说一下，我注意到这段代码返回了一个ar.length - 1个数的元素，所以如果你的数组有5个元素，新的打乱的数组将有4个元素。这是因为for循环说i>0。如果你改变为i>=0，你得到所有的元素洗牌。

使用随机类

  public static void randomizeArray(int[] arr) {

      Random rGenerator = new Random(); // Create an instance of the random class 
      for (int i =0; i< arr.length;i++ ) {
          //Swap the positions...

          int rPosition = rGenerator.nextInt(arr.length); // Generates an integer within the range (Any number from 0 - arr.length)
          int temp = arr[i]; // variable temp saves the value of the current array index;
          arr[i] = arr[rPosition];  // array at the current position (i) get the value of the random generated 
          arr[rPosition] = temp; // the array at the position of random generated gets the value of temp

      }

      for(int i = 0; i<arr.length; i++) {
          System.out.print(arr[i]); //Prints out the array
      } 

  }

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;
    }
}

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