当Fisher-Yates洗牌时Java堆栈中剩下的元素？

我在写一些代码时遇到了一些麻烦。基本上，我试图像一副扑克牌一样“洗牌”堆栈集合，但由于某种原因，我使用的临时堆栈中的一个不会完全清空，这会在下一次运行时导致一个空集合异常周围。我手动追踪代码和输出，并将元素留在临时堆栈1中（代码如下）。我真的不知道为什么会发生这种情况！如果你对此有所了解，这将会非常有帮助。当Fisher-Yates洗牌时Java堆栈中剩下的元素？

这是解决问题的方法链接：http://pastebin.com/cxJCmemZ

public void shuffleCards(LinkedStack<UnoCard> deck) { 
     int tempIndex; 
     LinkedStack<UnoCard> tempCardStack1 = new LinkedStack<UnoCard>(); 
     LinkedStack<UnoCard> tempCardStack2 = new LinkedStack<UnoCard>(); 

     //Fisher-Yates shuffle 
     for (int i = (deck.size() - 1); i >= 0; i--) { 
      tempIndex = ((int)(i * Math.random())); 

      System.out.println("i is: " + i); 
      System.out.println("tempIndex is: " + tempIndex); 

      //swap if cards are different 
      if (tempIndex != i) { 
       //pop face down cards up to first card onto temporary stack 
       System.out.println("Popping up to first card"); 
       for(int j = 0; j <= tempIndex; j++) { 
        UnoCard tempCard = faceDownCards.pop(); 
        System.out.println(tempCard.toString()); 
        tempCardStack1.push(tempCard); 
       } 

       //pop face down cards up to second card onto temporary stack 
       System.out.println("Popping up to second card"); 
       for(int j = (tempIndex + 1); j <= i; j++) { 
        UnoCard tempCard = faceDownCards.pop(); 
        System.out.println(tempCard.toString()); 
        tempCardStack2.push(tempCard); 
       } 

       //replace first card in second card position 
       System.out.println("Replacing first card"); 
       UnoCard tempCard = tempCardStack1.pop(); 
       System.out.println(tempCard.toString()); 
       faceDownCards.push(tempCard); 

       //place second card in temporary stack 
       System.out.println("Transferring second card"); 
       tempCard = tempCardStack2.pop(); 
       System.out.println(tempCard.toString()); 
       tempCardStack1.push(tempCard); 

       //replace temporary stack 
       System.out.println("Replacing second stack"); 
       for(int j = 0; j < tempCardStack2.size(); j++) { 
        tempCard = tempCardStack2.pop(); 
        System.out.println(tempCard.toString()); 
        faceDownCards.push(tempCard); 
       } 

       //replace second card in first card position 
       System.out.println("Replacing second card"); 
       tempCard = tempCardStack1.pop(); 
       System.out.println(tempCard.toString()); 
       faceDownCards.push(tempCard); 

       //replace temporary stack 
       System.out.println("Replacing first stack"); 
       for(int j = 0; j < tempCardStack1.size(); j++) { 
        tempCard = tempCardStack1.pop(); 
        System.out.println(tempCard.toString()); 
        faceDownCards.push(tempCard); 
       } 
      } 
     } 
    }

来源

2012-03-11 lollercopter

下一次你应该只是添加代码。并不是那么多。通常人们不希望遵循代码链接（特别是因为该链接可能会在未来消失，从而使问题和答案的价值较低）。 – AHungerArtist 2012-03-12 00:03:38

对不起，我是新来的。谢谢你纠正这个！ – lollercopter 2012-03-12 00:18:58

如果你弹出堆栈大小减小，因此for循环只会为大小/ 2次运行

因此结束循环应真的是

while(!tempCardStack1.isEmpty()){ 
    tempCard = tempCardStack1.pop(); 
    System.out.println(tempCard.toString()); 
    faceDownCards.push(tempCard); 
}

来源

2012-03-11 23:46:53

当Fisher-Yates洗牌时Java堆栈中剩下的元素？

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