我需要随机洗牌以下数组:

int[] solutionArray = {1, 2, 3, 4, 5, 6, 6, 5, 4, 3, 2, 1};

有什么函数可以做到吗?


当前回答

你现在可以使用java 8了:

Collections.addAll(list, arr);
Collections.shuffle(list);
cardsList.toArray(arr);

其他回答

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);
public class ShuffleArray {
public static void shuffleArray(int[] a) {
    int n = a.length;
    Random random = new Random();
    random.nextInt();
    for (int i = 0; i < n; i++) {
        int change = i + random.nextInt(n - i);
        swap(a, i, change);
    }
}

private static void swap(int[] a, int i, int change) {
    int helper = a[i];
    a[i] = a[change];
    a[change] = helper;
}

public static void main(String[] args) {
    int[] a = new int[] { 1, 2, 3, 4, 5, 6, 6, 5, 4, 3, 2, 1 };
    shuffleArray(a);
    for (int i : a) {
        System.out.println(i);
    }
}
}

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;
    }
}

使用Guava的int . aslist()就像这样简单:

Collections.shuffle(Ints.asList(array));