1195. Fizz Buzz Multithreaded
Description
You have the four functions:
printFizzthat prints the word"fizz"to the console,printBuzzthat prints the word"buzz"to the console,printFizzBuzzthat prints the word"fizzbuzz"to the console, andprintNumberthat 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"ifiis divisible by3and5,"fizz"ifiis divisible by3and not5,"buzz"ifiis divisible by5and not3, oriifiis not divisible by3or5.
Implement the FizzBuzz class:
FizzBuzz(int n)Initializes the object with the numbernthat represents the length of the sequence that should be printed.void fizz(printFizz)CallsprintFizzto output"fizz".void buzz(printBuzz)CallsprintBuzzto output"buzz".void fizzbuzz(printFizzBuzz)CallsprintFizzBuzzto output"fizzbuzz".void number(printNumber)Callsprintnumberto 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
Java
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();
}
}
}
}
C++
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++;
}
}
}
};