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。

无随机解:

   static void randomArrTimest(int[] some){
        long startTime = System.currentTimeMillis();
        for (int i = 0; i < some.length; i++) {
            long indexToSwap = startTime%(i+1);
            long tmp = some[(int) indexToSwap];
            some[(int) indexToSwap] = some[i];
            some[i] = (int) tmp;
        }
        System.out.println(Arrays.toString(some));
    }

这是knuth shuffle算法。

public class Knuth { 

    // this class should not be instantiated
    private Knuth() { }

    /**
     * Rearranges an array of objects in uniformly random order
     * (under the assumption that <tt>Math.random()</tt> generates independent
     * and uniformly distributed numbers between 0 and 1).
     * @param a the array to be shuffled
     */
    public static void shuffle(Object[] a) {
        int n = a.length;
        for (int i = 0; i < n; i++) {
            // choose index uniformly in [i, n-1]
            int r = i + (int) (Math.random() * (n - i));
            Object swap = a[r];
            a[r] = a[i];
            a[i] = swap;
        }
    }

    /**
     * Reads in a sequence of strings from standard input, shuffles
     * them, and prints out the results.
     */
    public static void main(String[] args) {

        // read in the data
        String[] a = StdIn.readAllStrings();

        // shuffle the array
        Knuth.shuffle(a);

        // print results.
        for (int i = 0; i < a.length; i++)
            StdOut.println(a[i]);
    }
}

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);
    }

}

Box from int[] to List<Integer> 用集合洗牌。洗牌法

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

List<Integer> list = Arrays.stream(solutionArray).boxed().collect(Collectors.toList());
Collections.shuffle(list);

System.out.println(list.toString());
// [1, 5, 5, 4, 2, 6, 1, 3, 3, 4, 2, 6]

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。