java - FizzBu​​zz 号码分组

Question

我正在尝试用 Java8 编写 FizzBu​​zz 问题。它工作正常，我得到了想要的输出。对于能被“3”整除的数字，它应该返回“Fizz”，对于能被“5”整除的数字，它应该返回“Buzz”，对于能被两者整除的数字，它应该返回“FizzBu​​zz”。

如果我将值作为“15”传递，它会返回:

["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。如果我将值作为“15”传递，我想获得如下输出:

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

我想按 Fizz、Buzz 和 FizzBu​​zz 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);
}
}

如果有人能指导我，我将不胜感激。谢谢

Answer 1

这里有一种使用流的方法:

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);
    }
}