原文链接: https://leetcode-cn.com/problems/fizz-buzz-multithreaded

英文原文

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

中文题目

编写一个可以从 1 到 n 输出代表这个数字的字符串的程序，但是：

• 如果这个数字可以被 3 整除，输出 "fizz"。
• 如果这个数字可以被 5 整除，输出 "buzz"。
• 如果这个数字可以同时被 3 和 5 整除，输出 "fizzbuzz"。

例如，当 n = 15，输出： 1, 2, fizz, 4, buzz, fizz, 7, 8, fizz, buzz, 11, fizz, 13, 14, fizzbuzz。

假设有这么一个类：

class FizzBuzz {
  public FizzBuzz(int n) { ... }               // constructor
  public void fizz(printFizz) { ... }          // only output "fizz"
  public void buzz(printBuzz) { ... }          // only output "buzz"
  public void fizzbuzz(printFizzBuzz) { ... }  // only output "fizzbuzz"
  public void number(printNumber) { ... }      // only output the numbers
}

请你实现一个有四个线程的多线程版  FizzBuzz， 同一个 FizzBuzz 实例会被如下四个线程使用：

1. 线程A将调用 fizz() 来判断是否能被 3 整除，如果可以，则输出 fizz。
2. 线程B将调用 buzz() 来判断是否能被 5 整除，如果可以，则输出 buzz。
3. 线程C将调用 fizzbuzz() 来判断是否同时能被 3 和 5 整除，如果可以，则输出 fizzbuzz。
4. 线程D将调用 number() 来实现输出既不能被 3 整除也不能被 5 整除的数字。

 

提示：

• 本题已经提供了打印字符串的相关方法，如 printFizz() 等，具体方法名请参考答题模板中的注释部分。

 

通过代码

高赞题解

这里其实就体现了Java并发包工具的方便。设立static（静态唯一)的CyclicBarrier（等待其他线程都一起触发之后，才进行下一步操作。）。

执行用时 :6 ms, 在所有 java 提交中击败了66.67%的用户
内存消耗 :36.3 MB, 在所有 java 提交中击败了100.00%的用户

private static CyclicBarrier barrier = new CyclicBarrier(4);

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

    // printFizz.run() outputs "fizz".
    public void fizz(Runnable printFizz) throws InterruptedException {
        for (int i = 1; i <= n; i++) {
            if (i % 3 == 0 && i % 5 != 0) {
                printFizz.run();
            }
            try {
                barrier.await();
            } catch (BrokenBarrierException e) {
                e.printStackTrace();
            }
        }
    }

    // printBuzz.run() outputs "buzz".
    public void buzz(Runnable printBuzz) throws InterruptedException {
        for (int i = 1; i <= n; i++) {
            if (i % 3 != 0 && i % 5 == 0) {
                printBuzz.run();
            }
            try {
                barrier.await();
            } catch (BrokenBarrierException e) {
                e.printStackTrace();
            }
        }
    }

    // printFizzBuzz.run() outputs "fizzbuzz".
    public void fizzbuzz(Runnable printFizzBuzz) throws InterruptedException {
        for (int i = 1; i <= n; i++) {
            if (i % 3 == 0 && i % 5 == 0) {
                printFizzBuzz.run();
            }
            try {
                barrier.await();
            } catch (BrokenBarrierException e) {
                e.printStackTrace();
            }
        }
    }

    // printNumber.accept(x) outputs "x", where x is an integer.
    public void number(IntConsumer printNumber) throws InterruptedException {
        for (int i = 1; i <= n; i++) {
            if (i % 3 != 0 && i % 5 != 0) {
                printNumber.accept(i);
            }
            try {
                barrier.await();
            } catch (BrokenBarrierException e) {
                e.printStackTrace();
            }
        }
    }
}

统计信息

通过次数	提交次数	AC比率
14518	22765	63.8%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言