Question

我正在写一些我正在写的代码有点麻烦。基本上，我试图像一副牌一样“洗牌”一个堆栈集合，但由于某种原因，我正在使用的一个临时堆栈不会完全清空，这会导致下一次运行时出现空集合异常周围。我手动跟踪代码和输出，并将元素留在临时堆栈1中（代码如下）。我真的不确定为什么会这样！如果您对此有任何见解，那将非常有用。

以下是问题方法的链接：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);
                }
            }
        }
    }

Answer 1

如果你弹出一个堆栈，那么大小会缩小，所以for循环只会运行大小/ 2次

所以结束循环应该是

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

当Fisher-Yates洗牌时，元素留在Java堆栈中？

1 个答案: