🌕 1195. 交替打印字符串

难度: 🌕

问题描述

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解法1 - Semaphore

class FizzBuzz {
    private int n;
    Semaphore s1;
    Semaphore s2;
    Semaphore s3;
    Semaphore s4;
    private int flag;

    public FizzBuzz(int n) {
        this.n = n;
        this.s1 = new Semaphore(0);
        this.s2 = new Semaphore(0);
        this.s3 = new Semaphore(0);
        this.s4 = new Semaphore(1);
    }

    // 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) {
                s1.acquire();
                printFizz.run();
                // System.out.println("fizz");
                int next = i + 1;
                if(next <= n) {
                    mySol(next);
                }
            }
        }
    }

    // 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) {
                s2.acquire();
                printBuzz.run();
                // System.out.println("buzz");
                int next = i + 1;
                if(next <= n) {
                    mySol(next);
                }                
            }
        }
    }

    // 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) {
                s3.acquire();
                printFizzBuzz.run();
                // System.out.println("fizzbuzz");
                int next = i + 1;
                if(next <= n) {
                    mySol(next);
                }                   
            }
        }
    }

    // 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) {
                s4.acquire();
                printNumber.accept(i);
                // System.out.println(i);
                // 判断接下来的数是哪种情况
                int next = i + 1;
                if(next <= n) {
                    mySol(next);
                }
            }
        }
    }

    private void mySol(int next) {
        if(next % 3 == 0 && next % 5 != 0) {
            s1.release();
        } else if(next % 3 != 0 && next % 5 == 0) {
            s2.release();
        } else if(next % 3 == 0 && next % 5 == 0) {
            s3.release();
        } else {
            s4.release();
        }
    }
}

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解法 2 - CountDownLatch

class FizzBuzz {
    // 思路：
    // 这种类似的题目，只要能用 Semaphore 做，就能用 CountDownLatch 做
    private int n;
    CountDownLatch c1;
    CountDownLatch c2;
    CountDownLatch c3;
    CountDownLatch c4;
    private int flag;

    public FizzBuzz(int n) {
        this.n = n;
        this.c1 = new CountDownLatch(1);
        this.c2 = new CountDownLatch(1);
        this.c3 = new CountDownLatch(1);
        this.c4 = new CountDownLatch(0);
    }

    // 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) {
                c1.await();
                printFizz.run();
                // System.out.println("fizz");
                int next = i + 1;
                if(next <= n) {
                    c1 = new CountDownLatch(1);
                    mySol(next);
                }
            }
        }
    }

    // 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) {
                c2.await();
                printBuzz.run();
                // System.out.println("buzz");
                int next = i + 1;
                if(next <= n) {
                    c2 = new CountDownLatch(1);
                    mySol(next);
                }                
            }
        }
    }

    // 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) {
                c3.await();
                printFizzBuzz.run();
                // System.out.println("fizzbuzz");
                int next = i + 1;
                if(next <= n) {
                    c3 = new CountDownLatch(1);
                    mySol(next);
                }                   
            }
        }
    }

    // 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) {
                c4.await();
                printNumber.accept(i);
                // System.out.println(i);
                // 判断接下来的数是哪种情况
                int next = i + 1;
                if(next <= n) {
                    c4 = new CountDownLatch(1);
                    mySol(next);
                }
            }
        }
    }

    private void mySol(int next) {
        if(next % 3 == 0 && next % 5 != 0) {
            c1.countDown();
        } else if(next % 3 != 0 && next % 5 == 0) {
            c2.countDown();
        } else if(next % 3 == 0 && next % 5 == 0) {
            c3.countDown();
        } else {
            c4.countDown();
        }
    }
}

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