如何阻止null打印

Question

执行我的程序时，它在第 13 行一直显示“null”我想知道我的算法出了什么问题，因为它一直打印 null。

private class SpadeIterator implements Iterator<Card>{
    private int nextCardSpade;
    private List<Card> cards;
    private int count=0;
    private SpadeIterator(List cards) {
        this.cards=cards;
        this.nextCardSpade = cards.size()-1;
    }

    @Override
    public boolean hasNext() {
        count++;
        if(nextCardSpade<0)
            return false;
        //nextCardSpade--;
        return true;
    }

   @Override
    public Card next() {

        int i=0;
        this.count=i;
        Card temp = cards.get(nextCardSpade);

        while(hasNext()){    //find SPADES
            temp=cards.get(nextCardSpade--);
            i++;

            if(temp.suit.value == Suit.SPADES.value)
                return temp;
        }
        //DONT MOVE
        return null;
        //nextCardSpade--;      //DONT DELETE

    }
}

Current Results

结果旨在显示 13 个黑桃，最后不返回 null。

Answer 1

检查 nextCardSpade 是否也等于 0：

if (nextCardSpade <= 0)

Answer 2

您的next() 方法不应包含任何会返回无效值的情况，例如null。如果没有要返回的下一个元素，则 hasNext() 方法的 job 返回 false 以防止您调用 next()。

所以你的代码应该看起来更像

class SpadeIterator implements Iterator<Card>{
    private int spadesCounter = 0;
    private Iterator<Card> cardsIt;

    private SpadeIterator(List<Card> cards) {
        cardsIt = cards.iterator();
    }

    @Override
    public boolean hasNext() {
        return spadesCounter<13; // we can't put spacesCounter++ here because 
                                 // we should be able to call `hasNext()` many times
                                 // and still get same answer,
                                 // so `hasNext()` shouldn't change any state 
                                 // (at least one which could cause changing its result)
    }

    @Override
    public Card next() {
        Card temp = cardsIt.next(); //if our `hasNext()` returned `false` but we 
                                    //didn't check or ignored it, this will CORRECTLY 
                                    //throw NoSuchElementException 
        while(temp.suit.value != Suit.SPADES.value){
            temp = cardsIt.next();
        }
        spadesCounter++;
        return temp;
    }
}

或者，如果您只是想遍历列表并仅打印选定的元素，您可以使用带有过滤功能的流

List<Card> cards = ...//not really important how get it
cards.stream()
     .filter(card -> card.suit.value == Suit.SPADES.value)
     .forEach(card -> System.out.println(card));

甚至更简单

for (Card card : cards){
    if(card.suit.value == Suit.SPADES.value){
        System.out.println(card);
    }
}