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1195. Fizz Buzz Multithreaded

Description

You have the four functions:

  • printFizz that prints the word "fizz" to the console,
  • printBuzz that prints the word "buzz" to the console,
  • printFizzBuzz that prints the word "fizzbuzz" to the console, and
  • printNumber that prints a given integer to the console.

You are given an instance of the class FizzBuzz that has four functions: fizz, buzz, fizzbuzz and number. The same instance of FizzBuzz will be passed to four different threads:

  • Thread A: calls fizz() that should output the word "fizz".
  • Thread B: calls buzz() that should output the word "buzz".
  • Thread C: calls fizzbuzz() that should output the word "fizzbuzz".
  • Thread D: calls number() that should only output the integers.

Modify the given class to output the series [1, 2, "fizz", 4, "buzz", ...] where the ith token (1-indexed) of the series is:

  • "fizzbuzz" if i is divisible by 3 and 5,
  • "fizz" if i is divisible by 3 and not 5,
  • "buzz" if i is divisible by 5 and not 3, or
  • i if i is not divisible by 3 or 5.

Implement the FizzBuzz class:

  • FizzBuzz(int n) Initializes the object with the number n that represents the length of the sequence that should be printed.
  • void fizz(printFizz) Calls printFizz to output "fizz".
  • void buzz(printBuzz) Calls printBuzz to output "buzz".
  • void fizzbuzz(printFizzBuzz) Calls printFizzBuzz to output "fizzbuzz".
  • void number(printNumber) Calls printnumber to output the numbers.

 

Example 1:

Input: n = 15
Output: [1,2,"fizz",4,"buzz","fizz",7,8,"fizz","buzz",11,"fizz",13,14,"fizzbuzz"]

Example 2:

Input: n = 5
Output: [1,2,"fizz",4,"buzz"]

 

Constraints:

  • 1 <= n <= 50

Solutions

Solution 1

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class FizzBuzz {
    private int n;

    public FizzBuzz(int n) {
        this.n = n;
    }

    private Semaphore fSema = new Semaphore(0);
    private Semaphore bSema = new Semaphore(0);
    private Semaphore fbSema = new Semaphore(0);
    private Semaphore nSema = new Semaphore(1);

    // printFizz.run() outputs "fizz".
    public void fizz(Runnable printFizz) throws InterruptedException {
        for (int i = 3; i <= n; i = i + 3) {
            if (i % 5 != 0) {
                fSema.acquire();
                printFizz.run();
                nSema.release();
            }
        }
    }

    // printBuzz.run() outputs "buzz".
    public void buzz(Runnable printBuzz) throws InterruptedException {
        for (int i = 5; i <= n; i = i + 5) {
            if (i % 3 != 0) {
                bSema.acquire();
                printBuzz.run();
                nSema.release();
            }
        }
    }

    // printFizzBuzz.run() outputs "fizzbuzz".
    public void fizzbuzz(Runnable printFizzBuzz) throws InterruptedException {
        for (int i = 15; i <= n; i = i + 15) {
            fbSema.acquire();
            printFizzBuzz.run();
            nSema.release();
        }
    }

    // printNumber.accept(x) outputs "x", where x is an integer.
    public void number(IntConsumer printNumber) throws InterruptedException {
        for (int i = 1; i <= n; i++) {
            nSema.acquire();
            if (i % 3 == 0 && i % 5 == 0) {
                fbSema.release();
            } else if (i % 3 == 0) {
                fSema.release();
            } else if (i % 5 == 0) {
                bSema.release();
            } else {
                printNumber.accept(i);
                nSema.release();
            }
        }
    }
}
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class FizzBuzz {
private:
    std::mutex mtx;
    atomic<int> index;
    int n;

    // 这里主要运用到了C++11中的RAII锁(lock_guard)的知识。
    // 需要强调的一点是,在进入循环后,要时刻不忘加入index <= n的逻辑
public:
    FizzBuzz(int n) {
        this->n = n;
        index = 1;
    }

    void fizz(function<void()> printFizz) {
        while (index <= n) {
            std::lock_guard<std::mutex> lk(mtx);
            if (