下面的代码将实现数组的随机排序。

// Shuffle the elements in the array
Collections.shuffle(Arrays.asList(array));

来自:http://www.programcreek.com/2012/02/java-method-to-shuffle-an-int-array-with-random-order/

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);
    }

}

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
      } 

  }

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);

其中一种解决方法是使用排列来预先计算所有的排列并存储在数组列表中

Java 8在Java .util. random类中引入了一个新方法ints()。ints()方法返回无限的伪随机int值流。您可以通过提供最小值和最大值来限制指定范围内的随机数。

Random genRandom = new Random();
int num = genRandom.nextInt(arr.length);

在生成随机数的帮助下，您可以遍历循环并与随机数交换当前索引。 这就是生成空间复杂度为O(1)的随机数的方法。

使用集合来洗牌一个基本类型数组有点过分……

你自己实现这个函数很简单，比如使用Fisher-Yates shuffle:

import java.util.*;
import java.util.concurrent.ThreadLocalRandom;

class Test
{
  public static void main(String args[])
  {
    int[] solutionArray = { 1, 2, 3, 4, 5, 6, 16, 15, 14, 13, 12, 11 };

    shuffleArray(solutionArray);
    for (int i = 0; i < solutionArray.length; i++)
    {
      System.out.print(solutionArray[i] + " ");
    }
    System.out.println();
  }

  // Implementing Fisher–Yates shuffle
  static void shuffleArray(int[] ar)
  {
    // If running on Java 6 or older, use `new Random()` on RHS here
    Random rnd = ThreadLocalRandom.current();
    for (int i = ar.length - 1; i > 0; i--)
    {
      int index = rnd.nextInt(i + 1);
      // Simple swap
      int a = ar[index];
      ar[index] = ar[i];
      ar[i] = a;
    }
  }
}