java：洗牌，

发布于 2024-10-20 14:34:00 字数 475 浏览 3 评论 0原文

这是一个面试问题。请给一些提示：

用向量实现一个方法，洗一副牌。

public class Card {
    private int value;
    Card(int v) {
        value = v;
        }

    public void print(){
        System.out.print(value+";");
    }
}



public class DeckShuffle {

    private final int num;
    Vector<Card> deck = new Vector<Card>();

// implement this shuffle function. DO NOT USE Collections.shuffle() !!
public void shuffle(){
// your code goes here!
}



}

原文

This is a interview-questions. Please give some hints:

Use vector to implement a method, shuffle a deck of Card.

public class Card {
    private int value;
    Card(int v) {
        value = v;
        }

    public void print(){
        System.out.print(value+";");
    }
}



public class DeckShuffle {

    private final int num;
    Vector<Card> deck = new Vector<Card>();

// implement this shuffle function. DO NOT USE Collections.shuffle() !!
public void shuffle(){
// your code goes here!
}



}

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信仰 2024-10-27 14:34:00

可以在 JDK 或 OpenJDK 的源包中找到，但算法非常简单（对于集合和数组来说基本上是相同的）：

given a[]
for i <- 0..a.length-2
  rnd_index <- random(i, a.length) #inclusive, exclusive
  swap a[i] and a[rnd_index]
next

Collections.shuffle() 的代码就位，因此您不需要额外的并行内存。它被称为Fisher Yates shuffle。

The code for Collections.shuffle() can be found in the source bundle for the JDK or from OpenJDK, but the algorithm is pretty simple (basically the same for a collection as for an array):

given a[]
for i <- 0..a.length-2
  rnd_index <- random(i, a.length) #inclusive, exclusive
  swap a[i] and a[rnd_index]
next

This works in place so you don't need extra parallel memory. It is known as the Fisher Yates shuffle.

回复收藏 0 原文

眼藏柔 2024-10-27 14:34:00

我想到的是：

public void shuffle() {
    Vector<Card> v = new Vector<Card>(deck.size());
    int size = deck.size();
    for(int i = 0; i < size; i++) {
        int index = Math.random() * deck.size();
        Card c = deck.remove(index);
        v.add(c);
    }
    deck = v;
}

这是干编码，没有进行测试。

Here's what comes to mind :

public void shuffle() {
    Vector<Card> v = new Vector<Card>(deck.size());
    int size = deck.size();
    for(int i = 0; i < size; i++) {
        int index = Math.random() * deck.size();
        Card c = deck.remove(index);
        v.add(c);
    }
    deck = v;
}

This is dry coded, no testing done.

回复收藏 0 原文

浮华 2024-10-27 14:34:00

void Shuffle()
{
  int n= deck.size();
  Vector<Card> v = new Vector<Card>(n);


  for (int i = 0; i < n; i++) {
     int j = (int)((i-1) * Math.random() )+ 1;
     if ( i != j ) {
         swap(cards, i, j);
     }
  }

  deck = v;
}

void Shuffle()
{
  int n= deck.size();
  Vector<Card> v = new Vector<Card>(n);


  for (int i = 0; i < n; i++) {
     int j = (int)((i-1) * Math.random() )+ 1;
     if ( i != j ) {
         swap(cards, i, j);
     }
  }

  deck = v;
}

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~没有更多了~