Writing FizzBuzz | 易学教程

问题

Reading the coding horror, I just came across the FizzBuzz another time.

The original post is here: Coding Horror: Why Can't Programmers.. Program?

For those who do not know: FizzBuzz is a quite popular childrens game. Counting from 1 to 100, and every time a number is divisible by 3 calling "Fizz", every time a number is divisible by 5 calling "Buzz" and every time a number is divisible by 3 and 5, calling "FizzBuzz instead of the number

But this time, I just started to code it down. It was a job of a minute, but there are several things that I do not like.

Here is my code

public void DoFizzBuzz()
{
    var combinations = new Tuple<int, string>[] 
    { 
        new Tuple<int, string> (3, "Fizz"), 
        new Tuple<int, string> (5, "Buzz"), 
    };

    for (int i = 1; i <= 100; ++i)
    {
        bool found = false;

        foreach (var comb in combinations)
        {
            if (i % comb.Item1 == 0)
            {
                found = true;
                Console.Write(comb.Item2);
            }
        }

        if (!found)
        {
            Console.Write(i);
        }

        Console.Write(Environment.NewLine);
    }
}

So my questions are:

How to get rid of the bool found?
Is there a better way of testing than the foreach?

回答1:

I think what you're trying to accomplish is a generic solution to FizzBuzz, that will work for any number of number-word combinations.

You have a good start - I think I can answer your questions with this example:

public void DoFizzBuzz()
{
    var combinations = new List<Tuple<int, string>>
    { 
        new Tuple<int, string> (3, "Fizz"), 
        new Tuple<int, string> (5, "Buzz"), 
    };

    Func<int, int, bool> isMatch = (i, comb) => i % comb == 0;
    for (int i = 1; i <= 100; i++)
    {
        Console.Write(i);

        var matchingCombs = combinations.Where(c => isMatch(i, c.Item1)).ToList();
        if (matchingCombs.Any())
        {
            Console.Write(string.Join("", matchingCombs.Select(c => c.Item2)));
        }
        else
        {
            Console.Write(i);
        }
        Console.Write(Environment.NewLine);
    }
}

In practice, you would pass combinations in to the method, but I included it inside just to be concise.

回答2:

I think your implementation is unnecessarily complex. This one does the job and is easier to understand:

public void DoFizzBuzz()
{
    for (int i = 1; i <= 100; i++)
    {
        bool fizz = i % 3 == 0;
        bool buzz = i % 5 == 0;
        if (fizz && buzz)
            Console.WriteLine ("FizzBuzz");
        else if (fizz)
            Console.WriteLine ("Fizz");
        else if (buzz)
            Console.WriteLine ("Buzz");
        else
            Console.WriteLine (i);
    }
}

回答3:

Take advantage of conditional format specifiers to get a nicely golfed version:

public void DoFizzBuzz()
{
    for(int i=1;i<101;i++)Console.WriteLine("{0:#;}{1:;;Fizz}{2:;;Buzz}",i%3*i%5==0?0:i,i%3,i%5);
}

回答4:

Unrolled for maximum efficiency. This program can outfizzbuzz all others.

public void FizzBuzz()
{
    const string FIZZ = "Fizz";
    const string BUZZ = "Buzz";
    const string FIZZBUZZ = "FizzBuzz";

    int i = 0;
    while (i < 150)
    {
        Console.WriteLine(++i);
        Console.WriteLine(++i);
        Console.WriteLine(FIZZ); ++i;
        Console.WriteLine(++i);
        Console.WriteLine(BUZZ); ++i;
        Console.WriteLine(FIZZ); ++i;
        Console.WriteLine(++i);
        Console.WriteLine(++i);
        Console.WriteLine(FIZZ); ++i;
        Console.WriteLine(BUZZ); ++i;
        Console.WriteLine(++i);
        Console.WriteLine(FIZZ); ++i;
        Console.WriteLine(++i);
        Console.WriteLine(++i);
        Console.WriteLine(FIZZBUZZ); ++i;
    }
}

回答5:

3rd edit:

Here is one way to "get rid of the bool" from your version (that is replace the for loop in your original question with this):

for (int i = 1; i <= 100; i++)
{
  var x = combinations.Where(n => i % n.Item1 == 0);

  if (x.Count() == 0)
    Console.Write(i);
  else
    Console.Write(string.Join("",x.Select(e => e.Item2)));

  Console.Write(Environment.NewLine);
}

Prior answers:

For a pure C# solution check out Keith Thompson's solution.

using System;
class FizzBuzz {
    static void Main() {
        for (int n = 1; n <= 100; n ++) {
            if (n % 15 == 0) {
                Console.WriteLine("FizzBuzz");
            }
            else if (n % 3 == 0) {
                Console.WriteLine("Fizz");
            }
            else if (n % 5 == 0) {
                Console.WriteLine("Buzz");
            }
            else {
                Console.WriteLine(n);
            }
        }
    }
}

I worked a bit on FixBuzz using linq. These are the solutions I came up with -- I believe they represent the best way to express the solution to this problem using Linq. (GitHub)

using System;
using System.Linq;

class FizzBuzz {
  static void Main() {
    var list = Enumerable.Range(1,100)
                .Select(n => {
                      if (n % 15 == 0) {
                        return "FizzBuzz";
                      }
                      if (n % 3 == 0) {
                        return "Fizz";
                      }
                      if (n % 5 == 0) {
                        return "Buzz";
                      }
                      return n.ToString();
                    });

    foreach(string item in list)
      Console.WriteLine(item);
  }
}

and the crazy one line version:

using System;
using System.Linq;

class FizzBuzz {
    static void Main() {
      Console.WriteLine(
      String.Join(
        Environment.NewLine,
        Enumerable.Range(1, 100)
          .Select(n => n % 15 == 0 ? "FizzBuzz" 
                     : n % 3 == 0 ? "Fizz" 
                     : n % 5 == 0 ? "Buzz" 
                     : n.ToString())
      ));
    }
}

回答6:

public void DoFizzBuzz()
{
    for (int i = 1; i <= 100; i++)
    {
        if (i % 3 == 0)
            Console.Write("Fizz");
        if (i % 5 == 0)
            Console.Write("Buzz");
        if (!(i % 3 == 0 || i % 5 == 0))
            Console.Write(i);

        Console.Write(Environment.NewLine);
    }
}

This gets rid of the bool found, but forces you to do duplicate evaluation. It is slightly different from some of the other answers using i % 15 == 0 for the FizzBuzz qualification. Whether or not this is better is up for debate. However, it is a different way.

回答7:

Did anyone do this one already?

Enumerable.Range(1, 100).Select(x =>
                (x % 15 == 0) ? "FIZZBUZZ"
                : (x % 5 == 0) ? "BUZZ"
                : (x % 3 == 0) ? "FIZZ"
                : x.ToString()
                )
                .ToList()
                .ForEach(console.WriteLine);

回答8:

Will add my 5 cents to solution by Linq. Everybody is using Select, which is basically Map function. IMHO foldl function suits better to solve this quiz:

Console.WriteLine(
                Enumerable
                .Range(1, 100)
                .Aggregate(new StringBuilder(), (builder, i)
                    => i % 15 == 0 ? builder.AppendLine("FizzBuzz")
                     : i % 3 == 0 ? builder.AppendLine("Fizz")
                     : i % 5 == 0 ? builder.AppendLine("Buzz")
                     : builder.AppendLine(i.ToString()))
                .ToString());

回答9:

Linq:

Enumerable.Range(1, 100).ToList().ForEach(i => Console.WriteLine( i % 3 * i % 5 == 0 ? (i % 3 == 0 ? "Fizz" : "") + (i % 5 == 0 ? "Buzz" : "") : i.ToString()));

回答10:

I think you started with a complicated way. Improving that code would be more complicated. You can use a temp variable to track and display that variable at the end of the FizzBuzz check. Below is code and you can also watch this detail c# FizzBuzz youtube video ( http://www.youtube.com/watch?v=OX5TM3q-JQg ) which explains how the below code is implemented.

    for (int j = 1; j <= 100; j++)
    {
    string Output = "";

    if (j % 3 == 0) Output = "Fizz";// Divisible by 3 --> Fizz

    if (j % 5 == 0) Output += "Buzz"; // Divisible by 5 --> Buzz

    if (Output == "") Output = j.ToString(); // If none then --> number

    Console.WriteLine(Output); // Finally print the complete output
    }

回答11:

Not the most efficient, but here's one using C#-6 string interpolation:

void Main()
{
    for (int i = 1; i <= 100; i++)
    {
       Console.WriteLine($"{(i % 15 == 0 ? "FizzBuzz" : 
                             i % 3 == 0 ? "Fizz" : 
                             i % 5 == 0 ? "Buzz" : i.ToString())}");
    }
}

回答12:

Enumerable.Range(1, 100).ToList().ForEach(i=>Console.WriteLine($"{(i%3*i%5==0?0:i):#;}{i%3:;;Fizz}{i%5:;;Buzz}"));

This answer has it all:

LINQ
Conditional formatting
String interpolation
All on a single line

Victory!

回答13:

Ok, what the heck, here's the solution I've come to like :)

public void DoFizzBuzz()
{
    for (int i = 1; i <= 100; ++i)
    {
        bool isDivisibleByThree = i % 3 == 0;
        bool isDivisibleByFive = i % 5 == 0;

        if (isDivisibleByThree || isDivisibleByFive)
        {
            if (isDivisibleByThree)
                cout << "Fizz";

            if (isDivisibleByFive)
                cout << "Buzz";
        }
        else
        {
            cout << i;
        }
        cout << endl;
    }
}

Obviously, this is not the fastest solution, but I like it because it emphasizes readability and makes the "FizzBuzz" case no longer a special case, but something that will happen naturally through the code path.

In the end, what I love most about this question whenever it comes up is that we get to see just how many different solutions ppl can come up with.

回答14:

You want probably make it configurable, but the question is what should be made configurable - we don't know that. Maybe we should make configurable all the cycle (FizzBuzz has the cycle). Here is very small and fun version with configurable cycle:

string[] fizzBuzzCycle = 
    "FizzBuzz,{0},{0},Fizz,{0},Buzz,Fizz,{0},{0},Fizz,Buzz,{0},Fizz,{0},{0}"
    .Split(',');

for (int i = 1; i <= 100; i++)
    Console.WriteLine(fizzBuzzCycle[i%fizzBuzzCycle.Length], i);

So if the strings or whole cycle should be changed it is easy to change. But you just don't know what to make configurable. Maybe condition will change: "for prime numbers print Pizz" and for this modification the solution by @ThomasLevesque is better, because it is easier to change.

回答15:

In my opinion, the FizzBuzz problem is always presented as a challenge to the interviwee to make the word FizzBuzz appear without explicitly printing it. Here is my solution in C#.

internal void PrintFizzBuzzAlternative(int num)
{
    if (num % 5 == 0)
        Console.Write("Fizz");
    if (num % 3 == 0)
        Console.Write("Buzz");
    if (num % 5 != 0 && num % 3 != 0)
        Console.Write(num);
    Console.WriteLine();
}

回答16:

The FizzBuzz question is a great interview question. We have started using it in our interview process. It is astounding how many people cannot solve such a simple problem.

Keep in mind, the original blog post was eventually locked due to a flood of people posting more solutions. Hahaha.

Regardless, here is mine in C++! ^_^

#include <iostream>
using namespace std;

int main(int argc, char** argv)
{
    for (int i = 1; i <= 100; ++i)
    {
        bool isMultipleOfThree = (i % 3) == 0;
        bool isMultipleOfFive = (i % 5) == 0;

        if (isMultipleOfThree) cout << "Fizz";
        if (isMultipleOfFive) cout << "Buzz";
        if (!isMultipleOfThree && !isMultipleOfFive) cout << i;

        cout << '\n';
    }

    return 0;
}

回答17:

With the input of Rob H and Jacob Krall here is what I have at the moment. Perhaps I will play around with that in future... just wanted to provide it.

public void DoFizzBuzz()
{
    // expect this to come in as parameter
    var combinations = new Tuple<int, string>[] 
    { 
        new Tuple<int, string> (3, "Fizz"), 
        new Tuple<int, string> (5, "Buzz"), 
    };

    Func<int, int, bool> isMatch = (i, comb) => i % comb == 0;

    // expect the borders 1, 100 to come in as parameters
    for (int i = 1; i <= 100; ++i)
    {
        var matchingCombs = combinations.Where(c => isMatch(i, c.Item1)).DefaultIfEmpty(new Tuple<int, string>(i, i.ToString())).Aggregate((v, w) => new Tuple<int, string>(v.Item1, v.Item2 + w.Item2)).Item2;
        Console.WriteLine(matchingCombs);
    }
}

回答18:

I would suggest this compact code as an addition to the previous simple and nice versions.

for (int i = 1; i <= 100; i++) // i++ but not ++i as in your example, be careful here
{
    bool fizz = i % 3 == 0;
    bool buzz = i % 5 == 0;
    string output = fizz && buzz ? "FizzBuzz" :
                            fizz ? "Fizz" :
                            buzz ? "Buzz" :
                            i.ToString();

    Console.WriteLn(output);
}

回答19:

I am a beginner, here is my attempt:

public void DoFizzBuzz()
   {
       for (int i = 1; i < 101; i++)
       {

           if ((i % 3 == 0) && (i % 5 == 0))
           {
               Console.WriteLine("{0} FizzBuzz", i);
           }
           else if (i % 3 == 0)
           {
               Console.WriteLine("{0} Fizz", i);
           }
           else if (i % 5 == 0)
           {
               Console.WriteLine("{0} Buzz", i);
           }
           else
           {
               Console.WriteLine(i);
           }

       }
       Console.ReadLine();
   }

Is there anything wrong with my approach? Mine seems a lot simpler than everyone's else approach so it must be wrong.

回答20:

A functional approach...

Console.WriteLine(Enumerable
    .Range(1,100)
    .Aggregate("", 
        (a,i) => a + "\n" + (i%15==0 ? "fizzbuzz" : 
                                (i%5==0 ? "buzz" :
                                    (i%3==0 ? "fizz" : i.ToString())))));

回答21:

I tried to solve this problem without looking at the answers. It took me 3 hours to succeed. (I'm just a hobby programmer by the way so don't bash me hard please :)) This is my c# version solution:

        static void Main(string[] args)
    {

        for (int i = 1; i <= 100; i++)
        {
            if(  ((i % 3) != 0) && ((i % 5) != 0))
            {
                WriteLine($"{i}");
            }
            else
            {
                if ((i % 15) == 0)
                {
                    WriteLine("FizzBuzz");
                }
                else if ((i % 3) == 0)
                {
                    WriteLine("Fizz");
                }
                else if ((i % 5) == 0)
                {
                    WriteLine("Buzz");
                }
            }                 
        }
    }

回答22:

Relatively simple solution using a for loop.

No Linq or anything - just basic shorthand if statements

for(int x=1;x<101;x++)
    Console.WriteLine(x%3==0?"Fizz"+(x%5==0?"Buzz":""):x%5==0?"Buzz":x+"");

The Linq solution which is a lot like csells (sans string interpolation) and fits on one line would be:

Enumerable.Range(1,100).ToList().ForEach(x=>Console.WriteLine(x%3==0?"Fizz"+(x%5==0?"Buzz":""):x%5==0?"Buzz":x+""));

回答23:

The null-coalescing operator is really useful:

string output = null;
for (int i = 1; i <= 100; i++)
{
     if (i % 3 == 0) output += "fizz";
     if (i % 5 == 0) output += "buzz";
     Console.WriteLine(output ?? i.ToString());
     output = null;
}
Console.ReadKey();

回答24:

I'll add mine even though there's 20 other solutions already written: It goes like this....

var x = 1;
while (x <= 100)
{
     if (x % 3 == 0 && x % 5 == 0)
        {Console.Writeline("FizzBuzz");}
     else if (x % 3 == 0)
        {Console.Writeline("fizz");}
     else if (x % 5 == 0)
        {Console.Writeline("Buzz");}
     else
        {Console.Writeline(x);}
     x++ 
}

First solution I came up with. Simple, to the point and gets the job done. No need for bool.

回答25:

I recommend using the ++i instead of the i++ in a for loop because i++ requires a copy to be made ;)

public void DoFizzBuzz()
{
    for (int i = 1; i < 101; ++i)
    {
        if (i % 15 == 0)
            Console.WriteLine ("FizzBuzz");
        else if (i % 3 == 0)
            Console.WriteLine ("Fizz");
        else if (i % 5 == 0)
            Console.WriteLine ("Buzz");
        else
            Console.WriteLine (i);
    }
}

回答26:

The original questions were: 1.How to get rid of the bool found? 2.Is there a better way of testing than the foreach?

This gets rid of the bool and the foreach, and I think it's still readable.

public static void DoFizzBuzz()
{
    var combinations = new Tuple<int, string>[]  
    {  
        new Tuple<int, string> (3, "Fizz"),  
        new Tuple<int, string> (5, "Buzz"),  
    };

    for (int i = 1; i <= 100; i++)
    {
        var fb = combinations.Where(t => {
            if (i % t.Item1 == 0)
            {
                Console.Write(t.Item2);
                return true;
            }
            return false;
        }).ToList();

        if (!fb.Any())
        {
            Console.Write(i);
        }

        Console.Write(Environment.NewLine);
    }
}

Who'd a thunk we'd be getting so excited about a simple kids game? :)

回答27:

You can use either use this and only take the amount you want

static void Main(string[] args)
{
    GetFizzBuzz().Take(100).ToList().ForEach(Console.WriteLine);
}

private static IEnumerable<string> GetFizzBuzz()
{
    for (var i = 0; i < int.MaxValue; i++)
    {
        if (i % 3 == 0 && i % 5 == 0) yield return "FizzBuzz";
        if (i % 3 == 0) yield return "Fizz";
        yield return i % 5 == 0 ? "Buzz" : i.ToString(CultureInfo.InvariantCulture);
    }
}

Or simply use this :

Enumerable.Range(1, 100).Select(s => {
    if (s % 3 == 0 && s % 5 == 0) return "FizzBuzz";
    if (s % 3 == 0) return "Fizz";
    return s%5 == 0 ? "Buzz" : s.ToString(CultureInfo.InvariantCulture);
}).ToList().ForEach(Console.WriteLine);

回答28:

Obviously this is a bit outside the spirit of the FizzBuzz challenge. But in my benchmark this was the fastest I could make it while single threaded and still terminating at 100. It is semi-unrolled and uses a StringBuilder. It is approximately three times faster than the standard approach.

const string FIZZ = " Fizz\n";
const string BUZZ = " Buzz\n";
const string FIZZBUZZ = " FizzBuzz\n";
    ...
var sb = new StringBuilder();
int i = 0;

while(true)
{       
    sb.Append(i+3);
    sb.Append(FIZZ);        
    sb.Append(i+5);
    sb.Append(BUZZ);        
    sb.Append(i+6);
    sb.Append(FIZZ);        
    sb.Append(i+9);
    sb.Append(FIZZ);        
    sb.Append(i+10);
    sb.Append(BUZZ);        
    if(i+12 > 100)
        break;
    sb.Append(i+12);
    sb.Append(FIZZ);    
    i+=15;
    sb.Append(i);
    sb.Append(FIZZBUZZ);
}

Console.Write(sb.ToString());

回答29:

With no if conditions, just one ternary operator.

string[] s = new string[6]{"Fizz", "Buzz", "", "", "", ""};
for (int i = 1; i <= 100; i++)
{
    string output = s[(i%3)*2] + s[(i%5)+1];
    Console.WriteLine(string.IsNullOrEmpty(output)? "" + i : output);
}

回答30:

Without using any If's, C# code.

 //False = 0, True = 1.
    private void DivisibilityByFiveThreeTest(int num)
    {
        string[,] values = new string [2,2]{
                             {"None","Fizz"},
                             {"Buzz","FizzBuzz"}
                             };
        for(int i=1;i< num;i++)
        Console.WriteLine(values[Convert.ToInt32(i % 5 == 0), Convert.ToInt32(i%3==0)]);

    }

来源：https://stackoverflow.com/questions/11764539/writing-fizzbuzz

标签

optimization

fizzbuzz