Collections类有一个有效的洗牌方法，可以复制，这样就不依赖于它:

/**
 * Usage:
 *    int[] array = {1, 2, 3};
 *    Util.shuffle(array);
 */
public class Util {

    private static Random random;

    /**
     * Code from method java.util.Collections.shuffle();
     */
    public static void shuffle(int[] array) {
        if (random == null) random = new Random();
        int count = array.length;
        for (int i = count; i > 1; i--) {
            swap(array, i - 1, random.nextInt(i));
        }
    }

    private static void swap(int[] array, int i, int j) {
        int temp = array[i];
        array[i] = array[j];
        array[j] = temp;
    }
}

下面是一个使用Apache Commons Math 3的解决方案。X(仅适用于int[]数组):

MathArrays.shuffle(array);

http://commons.apache.org/proper/commons-math/javadocs/api-3.6.1/org/apache/commons/math3/util/MathArrays.html shuffle (int [])

另外，Apache Commons Lang 3.6为ArrayUtils类引入了新的shuffle方法(用于对象和任何基本类型)。

ArrayUtils.shuffle(array);

http://commons.apache.org/proper/commons-lang/javadocs/api-release/org/apache/commons/lang3/ArrayUtils.html#shuffle-int:A-

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

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
      } 

  }

下面是使用集合的完整解决方案。洗牌的方法:

public static void shuffleArray(int[] array) {
  List<Integer> list = new ArrayList<>();
  for (int i : array) {
    list.add(i);
  }

  Collections.shuffle(list);

  for (int i = 0; i < list.size(); i++) {
    array[i] = list.get(i);
  }    
}

请注意，由于Java无法在int[]和Integer[]之间平滑转换(因此int[]和List<Integer>)，它受到了影响。

Collections类有一个有效的洗牌方法，可以复制，这样就不依赖于它:

/**
 * Usage:
 *    int[] array = {1, 2, 3};
 *    Util.shuffle(array);
 */
public class Util {

    private static Random random;

    /**
     * Code from method java.util.Collections.shuffle();
     */
    public static void shuffle(int[] array) {
        if (random == null) random = new Random();
        int count = array.length;
        for (int i = count; i > 1; i--) {
            swap(array, i - 1, random.nextInt(i));
        }
    }

    private static void swap(int[] array, int i, int j) {
        int temp = array[i];
        array[i] = array[j];
        array[j] = temp;
    }
}