这是另一种洗牌的方法

public List<Integer> shuffleArray(List<Integer> a) {
    List<Integer> b = new ArrayList<Integer>();
    while (a.size() != 0) {
        int arrayIndex = (int) (Math.random() * (a.size()));
        b.add(a.get(arrayIndex));
        a.remove(a.get(arrayIndex));
    }
    return b;
}

从原始列表中选择一个随机数，并将其保存在另一个列表中。然后从原始列表中删除该数字。原始列表的大小将继续减小1，直到所有元素都移动到新列表中。

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);

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

看看Collections类，特别是shuffle(…)。

使用Guava的int . aslist()就像这样简单:

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

import java.util.ArrayList;
import java.util.Random;
public class shuffle {
    public static void main(String[] args) {
        int a[] =  {1,2,3,4,5,6,7,8,9};
         ArrayList b = new ArrayList();
       int i=0,q=0;
       Random rand = new Random();

       while(a.length!=b.size())
       {
           int l = rand.nextInt(a.length);
//this is one option to that but has a flaw on 0
//           if(a[l] !=0)
//           {
//                b.add(a[l]);
//               a[l]=0;
//               
//           }
//           
// this works for every no. 
                if(!(b.contains(a[l])))
                {
                    b.add(a[l]);
                }



       }

//        for (int j = 0; j <b.size(); j++) {
//            System.out.println(b.get(j));
//            
//        }
System.out.println(b);
    }

}