Leetcode之多线程编程题

之多线程编程题"/>

Leetcode之多线程编程题

1116. 打印零与奇偶数

现有函数 printNumber 可以用一个整数参数调用，并输出该整数到控制台。

• 例如，调用 printNumber(7) 将会输出 7 到控制台。

给你类 ZeroEvenOdd 的一个实例，该类中有三个函数：zero、even 和 odd 。ZeroEvenOdd 的相同实例将会传递给三个不同线程：

• 线程 A：调用 zero() ，只输出 0
• 线程 B：调用 even() ，只输出偶数
• 线程 C：调用 odd() ，只输出奇数

修改给出的类，以输出序列 "010203040506..." ，其中序列的长度必须为 2n 。

实现 ZeroEvenOdd 类：

• ZeroEvenOdd(int n) 用数字 n 初始化对象，表示需要输出的数。
• void zero(printNumber) 调用 printNumber 以输出一个 0 。
• void even(printNumber) 调用printNumber 以输出偶数。
• void odd(printNumber) 调用 printNumber 以输出奇数。

Semaphore

class ZeroEvenOdd {private int n;private Semaphore zeroSema = new Semaphore(1);private Semaphore oddSema = new Semaphore(0);//奇数private Semaphore evenSema = new Semaphore(0);//偶数public ZeroEvenOdd(int n) {this.n = n;}public void zero(IntConsumer printNumber) throws InterruptedException {for (int i = 1; i <= n; i++) {zeroSema.acquire();printNumber.accept(0);if (i % 2!= 0) {//奇数oddSema.release();} else {evenSema.release();}}}public void even(IntConsumer printNumber) throws InterruptedException {for (int i = 1; i <= n; i++) {if (i % 2== 0) {//偶数 打印偶数 并释放zero的线程evenSema.acquire();printNumber.accept(i);zeroSema.release();}}}public void odd(IntConsumer printNumber) throws InterruptedException {for (int i = 1; i <= n; i++) {if (i % 2 != 0) {//奇数，打印奇数，并释放zero的线程oddSema.acquire();printNumber.accept(i);zeroSema.release();}}}
}

 synchronized

class ZeroEvenOdd {private int n;private final Object ob=new Object();private volatile int flag=0;public ZeroEvenOdd(int n) {this.n = n;}public void zero(IntConsumer printNumber) throws InterruptedException {for (int i = 1; i <= n; i++) {synchronized (ob){while (flag!=0){ob.wait();}printNumber.accept(0);if(i%2==0)flag=2;elseflag=1;ob.notifyAll();}}}public void even(IntConsumer printNumber) throws InterruptedException {for (int i = 1; i <= n; i++) {if (i % 2== 0) {//偶数 打印偶数 并释放zero的线程synchronized (ob){while (flag!=2){ob.wait();}printNumber.accept(i);flag=0;ob.notifyAll();}}}}public void odd(IntConsumer printNumber) throws InterruptedException {for (int i = 1; i <= n; i++) {if (i % 2!= 0) {//偶数 打印偶数 并释放zero的线程synchronized (ob){while (flag!=1){ob.wait();}printNumber.accept(i);flag=0;ob.notifyAll();}}}}
}

CountDownLatch

class ZeroEvenOdd {private int n;private CountDownLatch countDownLatch_zero=new CountDownLatch(0);private CountDownLatch countDownLatch_even=new CountDownLatch(1);private CountDownLatch countDownLatch_odd=new CountDownLatch(1);public ZeroEvenOdd(int n) {this.n = n;}public void zero(IntConsumer printNumber) throws InterruptedException {for (int i = 1; i <= n; i++) {countDownLatch_zero.await();printNumber.accept(0);countDownLatch_zero=new CountDownLatch(1);if (i % 2!= 0) {//奇数countDownLatch_odd.countDown();} else {countDownLatch_even.countDown();}}}public void even(IntConsumer printNumber) throws InterruptedException {for (int i = 1; i <= n; i++) {if (i % 2== 0) {//偶数 打印偶数 并释放zero的线程countDownLatch_even.await();printNumber.accept(i);countDownLatch_even=new CountDownLatch(1);countDownLatch_zero.countDown();}}}public void odd(IntConsumer printNumber) throws InterruptedException {for (int i = 1; i <= n; i++) {if (i % 2 != 0) {//奇数，打印奇数，并释放zero的线程countDownLatch_odd.await();printNumber.accept(i);countDownLatch_odd=new CountDownLatch(1);countDownLatch_zero.countDown();}}}
}

 Lock

class ZeroEvenOdd {private int n;private volatile int flag=0;Lock lock=new ReentrantLock();Condition condition_zero = lock.newCondition();Condition conditon_even = lock.newCondition();Condition condition_odd = lock.newCondition();public ZeroEvenOdd(int n) {this.n = n;}public void zero(IntConsumer printNumber) throws InterruptedException {for (int i = 1; i <= n; i++) {lock.lock();try{while (flag!=0){condition_zero.await();}printNumber.accept(0);if(i%2==0) {flag = 2;conditon_even.signal();}else {flag = 1;condition_odd.signal();}}finally {lock.unlock();}}}public void even(IntConsumer printNumber) throws InterruptedException {for (int i = 1; i <= n; i++) {if (i % 2== 0) {//偶数 打印偶数 并释放zero的线程lock.lock();try {while (flag != 2) {conditon_even.await();}printNumber.accept(i);flag = 0;condition_zero.signal();}finally {lock.unlock();}}}}public void odd(IntConsumer printNumber) throws InterruptedException {for (int i = 1; i <= n; i++) {if (i % 2!= 0) {//偶数 打印偶数 并释放zero的线程lock.lock();try {while (flag != 1) {condition_odd.await();}printNumber.accept(i);flag = 0;condition_zero.signal();}finally {lock.unlock();}}}}
}

Thread.yield

class ZeroEvenOdd {private int n;private volatile int flag=0;public ZeroEvenOdd(int n) {this.n = n;}public void zero(IntConsumer printNumber) throws InterruptedException {for (int i = 1; i <= n; i++) {while (flag!=0){Thread.yield();}printNumber.accept(0);if(i%2==0)flag=2;elseflag=1;}}public void even(IntConsumer printNumber) throws InterruptedException {for (int i = 1; i <= n; i++) {if (i % 2== 0) {//偶数 打印偶数 并释放zero的线程while (flag!=2){Thread.yield();}printNumber.accept(i);flag=0;}}}public void odd(IntConsumer printNumber) throws InterruptedException {for (int i = 1; i <= n; i++) {if (i % 2!= 0) {//偶数 打印偶数 并释放zero的线程while (flag!=1){Thread.yield();}printNumber.accept(i);flag=0;}}}
}

1195. 交替打印字符串

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) { ... }               // constructorpublic 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。

CyclicBarrier

class FizzBuzz {private int n;private CyclicBarrier cb = new CyclicBarrier(4);public FizzBuzz(int n) {this.n = n;}// printFizz.run() outputs "fizz".public void fizz(Runnable printFizz) throws InterruptedException, BrokenBarrierException {for (int i = 1; i <= n; i++) {if (i % 3 == 0 && i % 5 != 0) {printFizz.run();}cb.await();}}// printBuzz.run() outputs "buzz".public void buzz(Runnable printBuzz) throws InterruptedException, BrokenBarrierException {for (int i = 1; i <= n; i++) {if (i % 3 != 0 && i % 5 == 0) {printBuzz.run();}cb.await();}}// printFizzBuzz.run() outputs "fizzbuzz".public void fizzbuzz(Runnable printFizzBuzz) throws InterruptedException, BrokenBarrierException {for (int i = 1; i <= n; i++) {if (i % 3 == 0 && i % 5 == 0) {printFizzBuzz.run();}cb.await();}}// printNumber.accept(x) outputs "x", where x is an integer.public void number(IntConsumer printNumber) throws InterruptedException, BrokenBarrierException {for (int i = 1; i <= n; i++) {if (i % 3 != 0 && i % 5 != 0) {printNumber.accept(i);}cb.await();}}
}