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

最简单的洗牌代码:

import java.util.*;
public class ch {
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        ArrayList<Integer> l=new ArrayList<Integer>(10);
        for(int i=0;i<10;i++)
            l.add(sc.nextInt());
        Collections.shuffle(l);
        for(int j=0;j<10;j++)
            System.out.println(l.get(j));       
    }
}

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。

你现在可以使用java 8了:

Collections.addAll(list, arr);
Collections.shuffle(list);
cardsList.toArray(arr);

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
      } 

  }
Random rnd = new Random();
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;
}

顺便说一下,我注意到这段代码返回了一个ar.length - 1个数的元素,所以如果你的数组有5个元素,新的打乱的数组将有4个元素。这是因为for循环说i>0。如果你改变为i>=0,你得到所有的元素洗牌。