【发布时间】:2020-01-16 04:17:51
【问题描述】:
基本上我有这个 toString 方法女巫显示排序的卡片。基本上,如果我有一副完整的牌,这就是输出的样子。
{Ace of Clubs | Two of Clubs | Three of Clubs | Four of Clubs | Five of Clubs | Six of Clubs | Seven of Clubs | Eight of Clubs | Nine of Clubs | Ten of Clubs | Jack of Clubs | Queen of Clubs | King of Clubs | Ace of Diamonds | Two of Diamonds | Three of Diamonds | Four of Diamonds | Five of Diamonds | Six of Diamonds | Seven of Diamonds | Eight of Diamonds | Nine of Diamonds | Ten of Diamonds | Jack of Diamonds | Queen of Diamonds | King of Diamonds | Ace of Hearts | Two of Hearts | Three of Hearts | Four of Hearts | Five of Hearts | Six of Hearts | Seven of Hearts | Eight of Hearts | Nine of Hearts | Ten of Hearts | Jack of Hearts | Queen of Hearts | King of Hearts | Ace of Spades | Two of Spades | Three of Spades | Four of Spades | Five of Spades | Six of Spades | Seven of Spades | Eight of Spades | Nine of Spades | Ten of Spades | Jack of Spades | Queen of Spades | King of Spades}
我目前通过一个简单的循环来完成此操作,该循环在开始处放置一个花括号,并循环显示卡片组中有多少元素,添加 deck.get(i).getCard() +“|”的输出。最后从末尾减去三个字符并放置另一个花括号。但这不是一个非常花哨的解决方案,而且非常程序化,我喜欢优雅的解决方案,我认为存在用于这种目的的函数,但我真的不知道 lambda 表达式或其中任何一个,所以我想知道是否有人可以走路我通过我怎么能做到这一点。在下面,您会找到我当前的解决方案。任何帮助表示赞赏干杯!
public String toString(){
Collections.sort(deck);
String result = "{";
for(int i = 0; i < deck.size(); i++){
result += deck.get(i).getCard() + " | ";
}
if(deck.size() == 0){
return "{}";
}
result = result.substring(0, result.length() - 3);
result += "}";
shuffle();
return result;
}
【问题讨论】:
-
可以说,Java 8 中最被低估的新增功能之一,
StringJoiner -
我会更担心
sort和shuffle。只是重新安排甲板打印出来,呃。如果您不能提前使用已排序的卡片组,请至少制作一份副本并对其进行排序。
标签: java lambda expression