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

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);
    }

}

其他回答

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;
    }
  }
}

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));
    }

下面是数组的泛型版本:

import java.util.Random;

public class Shuffle<T> {

    private final Random rnd;

    public Shuffle() {
        rnd = new Random();
    }

    /**
     * Fisher–Yates shuffle.
     */
    public void shuffle(T[] ar) {
        for (int i = ar.length - 1; i > 0; i--) {
            int index = rnd.nextInt(i + 1);
            T a = ar[index];
            ar[index] = ar[i];
            ar[i] = a;
        }
    }
}

考虑到ArrayList基本上只是一个数组,使用ArrayList而不是显式数组并使用Collections.shuffle()可能是明智的。但是,性能测试并没有显示上述方法与Collections.sort()之间有任何显著差异:

Shuffe<Integer>.shuffle(...) performance: 576084 shuffles per second
Collections.shuffle(ArrayList<Integer>) performance: 629400 shuffles per second
MathArrays.shuffle(int[]) performance: 53062 shuffles per second

Apache Commons实现MathArrays。Shuffle限制为int[],性能损失可能是由于使用了随机数生成器。

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
      } 

  }

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。