那么有没有办法简化这个以使其更小?
else if(selectedCards.size() == 3
&& cardAt(selectedCards.get(0)).pointValue() + cardAt(selectedCards.get(1)).pointValue() + cardAt(selectedCards.get(2)).pointValue() == 0
&& !cardAt(selectedCards.get(0)).rank().equals(cardAt(selectedCards.get(1)).rank())
&& !cardAt(selectedCards.get(0)).rank().equals(cardAt(selectedCards.get(2)).rank())
&& !cardAt(selectedCards.get(1)).rank().equals(cardAt(selectedCards.get(2)).rank()))
答案 0 :(得分:3)
从我可以看到你试图测试3张牌是否有不同的等级。一种更简单的测试方法是将它们放入一个Set中,看看设置的大小是否与选定的set相同。这可以扩展到任意数量的选定卡片......
public boolean differentRanks(List<Integer> selectedCards) {
Set<Integer> ranks = new HashSet<Integer>();
for (int card : selectedCards) {
ranks.add(cardAt(card).rank());
}
return ranks.size() == selectedCards.size();
}
我还会创建一种方法来累计所选卡片的分数......
public int sum(List<Integer> selectedCards) {
int total;
for (int card : selectedCards) {
total += cardAt(card).pointValue();
}
return total;
}
所以条件会结束
} else if (selectedCards.size() == 3 && sum(selectedCards) == 0 &&
differentRanks(selectedCards) {
答案 1 :(得分:0)
这是一个选择:
else if(selectedCards.size() == 3
&& cardAt(selectedCards.get(0)).pointValue() + cardAt(selectedCards.get(1)).pointValue() + cardAt(selectedCards.get(2)).pointValue() == 0
&& !(cardAt(selectedCards.get(0)).rank().equals(cardAt(selectedCards.get(1)).rank())).equals(cardAt(selectedCards.get(2)).rank()) )
为了使这种情况更具可读性,你可以这样做:
//here you extract the values you need only once and use them in your condition block below
int cardsSize = selectedCards.size();
int pointValue0 = cardsSize == 3 ? cardAt(selectedCards.get(0)).pointValue() : 0;
int pointValue1 = cardsSize == 3 ? cardAt(selectedCards.get(1)).pointValue() : 0;
int pointValue2 = cardsSize == 3 ? cardAt(selectedCards.get(2)).pointValue() : 0;
bool rankEquals = CompareRanks(cardAt(selectedCards.get(0)),cardAt(selectedCards.get(1)),cardAt(selectedCards.get(2));
if (<condition>) {
//block of sentences
} else if (cardsSize == 3 && (pointValue0 + pointValue1 + pointValue2) == 0 && !rankEquals )
我建议创建一个名为&#34; CompareRanks&#34;你收到3个不同的对象(&#34; cardAt&#34;的结果)并在那里得到排名并比较值是否相同。
此选项可引导您使用更多代码行,但除了您之外的任何人都更清晰,更易读。
答案 2 :(得分:0)
在我看来最具可读性:
else if(selectedCards.size() == 3 && checkRanks(selectedCards))
{
//...
}
//...
private boolean checkRanks(List<Card> cards)
{
Card zeroCard = cardAt(selectedCards.get(0));
Card firstCard = cardAt(selectedCards.get(1));
Card secondCard = cardAt(selectedCards.get(2));
boolean isZero = zeroCard.pointValue() + firstCard.pointValue() + secondCard.pointValue() == 0;
boolean zeroCardRankNotEqualFirst = !zeroCard.rank().equals(firstCard.rank())
boolean zeroCardRankNotEqualSecond = !zeroCard.rank().equals(secondCard.rank())
boolean firstCardRankNotEqualsSecond = !firstCard.rank().equals(secondCard.rank());
return isZero && zeroCardRankNotEqualFirst && zeroCardRankNotEqualSecond && firstCardRankNotEqualsSecond;
}