FizzBuzz数字分组

Question

我正试图用Java8编写FizzBuzz问题的代码。它运行良好，我得到了期望的输出。对于可被“3”整除的数字，它应该返回“Fizz”；对于可被“5”整除的数字，它应该返回“Buzz”；对于可被两者整除的数字，它应该返回“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。我想得到如下输出，如果我将值传递为“15”：

{"Fizz": [3, 6, 9, 12],"Buzz": [5, 10],"FizzBuzz": [15]}

我想把数字按嘶嘶声、嗡嗡声和嘶嘶声进行分组。

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);
}
}

如果有人能给我指路我将感激不尽谢谢

Answer 1

Map<String, List<Integer>> map = 
  IntStream.rangeClosed(1, 15)
    .mapToObj(n -> {
        String str = n % 15 == 0 ? "FizzBuzz" : 
            (n % 5 == 0) ? "Buzz" : (n % 3 == 0) ? "Fizz" : null;
        return str == null ? null : new SimpleEntry<>(str, n);
    }).filter(Objects::nonNull)
    .collect(groupingBy(Entry::getKey, mapping(Entry::getValue, toList())));

System.out.println(map);

Answer 2

我会返回地图

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;
}

Answer 3

以下是一种使用流的方法：

Map<String, List<String>> result = IntStream.rangeClosed(1, n)
    .filter(i -> i % 3 == 0 || i % 5 == 0)
    .boxed()
    .collect(Collectors.groupingBy(i ->
        i % 15 == 0 ? "FizzBuzz" :
        i %  3 == 0 ? "Fizz" : 
                      "Buzz"));

没有溪流：

Map<String, List<String>> result = new HashMap<>();
for (int i = 0; i < n; i++) {
    if (i % 3 == 0 || i % 5 == 0) {
        String key = i % 15 == 0 ? "FizzBuzz" :
                     i %  3 == 0 ? "Fizz" : 
                                   "Buzz";
        result.computeIfAbsent(key, k -> new ArrayList<>()).add(i);
    }
}