Java FizzBuzz 1 行

Question

如何仅用 return 语句完成 Codingbat 中的 FizzBuzz 练习？

我上次使用的解决问题的代码是：

public String[] fizzBuzz(int start, int end) { String[] a = new String[end - start]; for(int i = start; i < end; i++) a[i - start] = i % 15 == 0 ? "FizzBuzz" : i % 3 == 0 ? "Fizz" : i % 5 == 0 ? "Buzz" : String.valueOf(i); return a; }

我的目标是让代码看起来像这样：

public String[] fizzBuzz(int start, int end) { return foo; }

问题

This is slightly more difficult version of the famous FizzBuzz problem which is sometimes given as a first problem for job interviews. (See also: FizzBuzz Code.) Consider the series of numbers beginning at start and running up to but not including end, so for example start=1 and end=5 gives the series 1, 2, 3, 4. Return a new String[] array containing the string form of these numbers, except for multiples of 3, use "Fizz" instead of the number, for multiples of 5 use "Buzz", and for multiples of both 3 and 5 use "FizzBuzz". In Java, String.valueOf(xxx) will make the String form of an int or other type. This version is a little more complicated than the usual version since you have to allocate and index into an array instead of just printing, and we vary the start/end instead of just always doing 1..100.

测试用例

fizzBuzz(1, 6) → ["1", "2", "Fizz", "4", "Buzz"] fizzBuzz(1, 8) → ["1", "2", "Fizz", "4", "Buzz", "Fizz", "7"] fizzBuzz(1, 11) → ["1", "2", "Fizz", "4", "Buzz", "Fizz", "7", "8", "Fizz", "Buzz"] fizzBuzz(1, 16) → ["1", "2", "Fizz", "4", "Buzz", "Fizz", "7", "8", "Fizz", "Buzz", "11", "Fizz", "13", "14", "FizzBuzz"] fizzBuzz(1, 4) → ["1", "2", "Fizz"] fizzBuzz(1, 2) → ["1"] fizzBuzz(50, 56) → ["Buzz", "Fizz", "52", "53", "Fizz", "Buzz"] fizzBuzz(15, 17) → ["FizzBuzz", "16"] fizzBuzz(30, 36) → ["FizzBuzz", "31", "32", "Fizz", "34", "Buzz"] fizzBuzz(1000, 1006) → ["Buzz", "1001", "Fizz", "1003", "1004", "FizzBuzz"] fizzBuzz(99, 102) → ["Fizz", "Buzz", "101"] fizzBuzz(14, 20) → ["14", "FizzBuzz", "16", "17", "Fizz", "19"]

Answer 1

不确定为什么要在单个 return 语句中执行此操作，但尽管如此：

return IntStream.range(start, end)
                .mapToObj(i -> i % 15 == 0 ? "FizzBuzz" : 
                          i % 3 == 0 ? "Fizz" : i % 5 == 0 ? "Buzz" : String.valueOf(i))
                .toArray(String[]::new);

IntStream.range returns 从 start（含）到 end（不含）的顺序 IntStream，递增步长为 1，正好符合以您命令式的方式。
mapToObj 使我们能够将 IntStream 转换为 Stream<String>.
toArray 使我们能够将 Stream<String> 转换为 String[] for return.

您可以阅读有关 IntStream API here 的更多信息。

Answer 2

确保你有 JDK 8（否则此代码将无法运行）
使用IntStream.range你可以生成整数。一旦你有了一堆数字，你就必须相应地映射它们。（在这种情况下，我们必须将它们映射到 String [] 并检查多个 3、多个 5 以及多个 3 & 5）现在，对于其余的数字，我们可以使用 String.valueof() 将它们添加到数组中。

N.B：确保将 FIZZBUZZ 放在顶部，因为它是一个 if 语句，只有其中一个被执行。

    public static String[] foo(int start, int end) {
        return IntStream.range(start, end).mapToObj(i-> ( 
        i%5==0 && i%3==0 ) ? "FIZZBUZZ": 
        i%3==0 ? "FIZZ":
        i%5==0 ? "BUZZ": 
        String.valueOf(i)).toArray(String[]::new);
}

Answer 3

在没有任何 Java8 功能的情况下，我能想到的唯一方法涉及一些丑陋的数组到字符串的转换，以及字符串数组的字符串到字符串数组的一些非常...实用的正则表达式。但我想这个解决方案无论如何都不应该是完美的。但是，它通过了所有测试用例，因此它实现了测试驱动开发可以实现的所有目标...

永远不要在现实生活中写任何类似的东西！

import java.util.Arrays;

public class OneLineFizzBuzz
{
    public static void main(String[] args)
    {
        OneLineFizzBuzz f = new OneLineFizzBuzz();
        String[] result = f.fizzBuzz(1000, 1006);
        System.out.println(Arrays.toString(result));

        check(f.fizzBuzz(1, 6), new String[]{"1", "2", "Fizz", "4", "Buzz"});
        check(f.fizzBuzz(1, 8), new String[]{"1", "2", "Fizz", "4", "Buzz", "Fizz", "7"});
        check(f.fizzBuzz(1, 11), new String[]{"1", "2", "Fizz", "4", "Buzz", "Fizz", "7", "8", "Fizz", "Buzz"});
        check(f.fizzBuzz(1, 16), new String[]{"1", "2", "Fizz", "4", "Buzz", "Fizz", "7", "8", "Fizz", "Buzz", "11", "Fizz", "13", "14", "FizzBuzz"});
        check(f.fizzBuzz(1, 4), new String[]{"1", "2", "Fizz"});
        check(f.fizzBuzz(1, 2), new String[]{"1"});
        check(f.fizzBuzz(50, 56), new String[]{"Buzz", "Fizz", "52", "53", "Fizz", "Buzz"});
        check(f.fizzBuzz(15, 17), new String[]{"FizzBuzz", "16"});
        check(f.fizzBuzz(30, 36), new String[]{"FizzBuzz", "31", "32", "Fizz", "34", "Buzz"});
        check(f.fizzBuzz(1000, 1006), new String[]{"Buzz", "1001", "Fizz", "1003", "1004", "FizzBuzz"});
        check(f.fizzBuzz(99, 102), new String[]{"Fizz", "Buzz", "101"});
        check(f.fizzBuzz(14, 20), new String[]{"14", "FizzBuzz", "16", "17", "Fizz", "19"});        
    }
    private static void check(String[] fizzBuzz, String[] strings)
    {
        boolean passed = Arrays.equals(fizzBuzz, strings);
        System.out.println(Arrays.toString(fizzBuzz) + " passed? " + passed);
        if (!passed)
        {
            System.out.println("Whoopsie...");
        }
    }

    public String[] fizzBuzz(int start, int end)
    {
        return (start == end ? ""
            : (start % 15 == 0 ? "FizzBuzz"
                : start % 3 == 0 ? "Fizz"
                    : start % 5 == 0 ? "Buzz" : String.valueOf(start))
                + Arrays.toString(fizzBuzz(start + 1, end)))
                    .replaceAll("\]", "").split("\[|,\s*");
    }
}

Java FizzBuzz 1 行

Java FizzBuzz 1 Line

java

fizzbuzz

问题

测试用例

永远不要在现实生活中写任何类似的东西！

Java FizzBu​​zz 1 行

Java FizzBuzz 1 Line

java

fizzbuzz

问题

测试用例

永远不要在现实生活中写任何类似的东西！

Java FizzBuzz 1 行