LeetCode 1195. Fizz Buzz Multithreaded Solution in Java & C++ | Explanation + Code

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

JavaC++

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(); } } } }(code-box)

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 (0 == index % 3 && 0 != index % 5 && index <= n) { printFizz(); index++; } } } void buzz(function<void()> printBuzz) { while (index <= n) { std::lock_guard<std::mutex> lk(mtx); if (0 == index % 5 && 0 != index % 3 && index <= n) { printBuzz(); index++; } } } void fizzbuzz(function<void()> printFizzBuzz) { while (index <= n) { std::lock_guard<std::mutex> lk(mtx); if (0 == index % 15 && index <= n) { printFizzBuzz(); index++; } } } void number(function<void(int)> printNumber) { while (index <= n) { std::lock_guard<std::mutex> lk(mtx); if (0 != index % 3 && 0 != index % 5 && index <= n) { printNumber(index); index++; } } } };(code-box)