我正在尝试在 Java8 中编写 FizzBuzz 问题。它工作正常,我得到了想要的输出。对于可被“3”整除的数,应返回“Fizz”,对于可被“5”整除的数,应返回“Buzz”,对于可被两者整除的数,应返回“FizzBuzz”。
如果我将值传递为“15”,它将返回:
["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"]
现在,我被困在一件事上。如果我将值传递为“15”,我想获得如下输出:
{"Fizz": [3, 6, 9, 12],"Buzz": [5, 10],"FizzBuzz": [15]}
我想按 Fizz、Buzz 和 FizzBuzz 对数字进行分组。
这是我的代码:
public class FizzBuzzService {
private Map<Rule, String> ruleContainers = new HashMap();
private Set<Rule> rules = new HashSet();
public FizzBuzzService(){
addRule(i -> i % 15 == 0, "FizzBuzz");
addRule(i -> i % 3 == 0, "Fizz");
addRule(i -> i % 5 == 0, "Buzz");
}
public void addRule(Rule rule, String res) {
rules.add(rule);
ruleContainers.put(rule, res);
}
public String getValue(int i) {
for (Rule rule : rules) {
if (rule.apply(i)) {
return ruleContainers.get(rule);
}
}
return String.valueOf(i);
}
//then the origin code should be as follows:
public List<String> fizzBuzz(int n) {
List<String> res = new ArrayList();
for(int i = 1; i <= n; i++){
res.add(getValue(i));
}
return res;
}
interface Rule{
boolean apply(int i);
}
}
如果有人可以指导我,我将不胜感激。谢谢
最佳答案
我会返回一个 Map<String, List<Integer>>
(使用 LinkedHashMap
而不是 HashMap
并且您的键将保留插入顺序)为 List<Integer>
、 fizz
和 buzz
值创建 fizzbuzz
(s)。它可以在单个 static
方法中完成,例如,
public static Map<String, List<Integer>> fizzBuzz(int n) {
Map<String, List<Integer>> map = new HashMap<>();
List<Integer> fizz = new ArrayList<>(), //
buzz = new ArrayList<>(), //
fizzbuzz = new ArrayList<>();
IntStream.rangeClosed(1, n).forEachOrdered(i -> {
boolean f = i % 3 == 0, b = i % 5 == 0;
if (f && b) {
fizzbuzz.add(i);
} else if (f) {
fizz.add(i);
} else if (b) {
buzz.add(i);
}
});
map.put("Fizz", fizz);
map.put("Buzz", buzz);
map.put("FizzBuzz", fizzbuzz);
return map;
}
关于java - FizzBuzz 数字的分组,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/48354492/