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);

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);

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。

在Java中，我们可以使用集合。Shuffle方法随机重新排列列表中的项目。

Groovy 3.0.0直接将shuffle和shuffle方法添加到List或数组中。

最简单的解决方案，这种随机洗牌数组。

String location[] = {"delhi","banglore","mathura","lucknow","chandigarh","mumbai"};
int index;
String temp;
Random random = new Random();
for(int i=1;i<location.length;i++)
{
    index = random.nextInt(i+1);
    temp = location[index];
    location[index] = location[i];
    location[i] = temp;
    System.out.println("Location Based On Random Values :"+location[i]);
}