leetcode 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) { ... } // 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:
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。
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。
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。
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。
解法零: 什么都不需要,我们直接while循环来反复判断条件是否成立,问题是由于num不是volatile的,不能及时更新,很多线程都在无意义的死循环(超出时间限制)
class FizzBuzz {
private int n;
private int num =1;
public FizzBuzz(int n) {
this.n = n;
}
// printFizz.run() outputs "fizz".
public void fizz(Runnable printFizz) throws InterruptedException {
while(num<=n){
if(num%3==0 && num%5!=0){
printFizz.run();
num++;
}
}
}
// printBuzz.run() outputs "buzz".
public void buzz(Runnable printBuzz) throws InterruptedException {
while(num<=n){
if(num%3!=0 && num%5==0){
printBuzz.run();
num++;
}
}
}
// printFizzBuzz.run() outputs "fizzbuzz".
public void fizzbuzz(Runnable printFizzBuzz) throws InterruptedException {
while(num<=n){
if(num%3==0 && num%5==0){
printFizzBuzz.run();
num++;
}
}
}
// printNumber.accept(x) outputs "x", where x is an integer.
public void number(IntConsumer printNumber) throws InterruptedException {
while(num<=n){
if(num%3!=0 && num%5!=0){
printNumber.accept(num);
num++;
}
}
}
}
解法一:volatile (因为同一时刻可能有多个线程在读取变量,但是必然只有一个线程在写入变量,所以使用volatile完全没有问题)
class FizzBuzz {
private int n;
private volatile int num = 1;
public FizzBuzz(int n) {
this.n = n;
}
// printFizz.run() outputs "fizz".
public void fizz(Runnable printFizz) throws InterruptedException {
while(num<=n){
if(num%3==0 && num%5!=0){
printFizz.run();
num++;
}
}
}
// printBuzz.run() outputs "buzz".
public void buzz(Runnable printBuzz) throws InterruptedException {
while(num<=n){
if(num%3!=0 && num%5==0){
printBuzz.run();
num++;
}
}
}
// printFizzBuzz.run() outputs "fizzbuzz".
public void fizzbuzz(Runnable printFizzBuzz) throws InterruptedException {
while(num<=n){
if(num%3==0 && num%5==0){
printFizzBuzz.run();
num++;
}
}
}
// printNumber.accept(x) outputs "x", where x is an integer.
public void number(IntConsumer printNumber) throws InterruptedException {
while(num<=n){
if(num%3!=0 && num%5!=0){
printNumber.accept(num);
num++;
}
}
}
}
解法二:既然volatile是可以的,那么AtomicInteger原子操作类必然也是可以的(代码几乎完全一致这里就不再演示了)
解法三:synchronized + volatile (我们知道在某一时刻,必然只有一个线程的操作是有意义的,所以我们可以引入synchronized来让其它线程wait(),到时候在notify)
class FizzBuzz {
private int n;
private volatile int num =1;
private Object lock = new Object();
public FizzBuzz(int n) {
this.n = n;
}
// printFizz.run() outputs "fizz".
public void fizz(Runnable printFizz) throws InterruptedException {
synchronized(lock){
while(num<=n){
if(num%3==0 && num%5!=0){
printFizz.run();
num++;
lock.notifyAll();
}else{
lock.wait();
}
}
}
}
// printBuzz.run() outputs "buzz".
public void buzz(Runnable printBuzz) throws InterruptedException {
synchronized(lock){
while(num<=n){
if(num%3!=0 && num%5==0){
printBuzz.run();
num++;
lock.notifyAll();
}else{
lock.wait();
}
}
}
}
// printFizzBuzz.run() outputs "fizzbuzz".
public void fizzbuzz(Runnable printFizzBuzz) throws InterruptedException {
synchronized(lock){
while(num<=n){
if(num%3==0 && num%5==0){
printFizzBuzz.run();
num++;
lock.notifyAll();
}else{
lock.wait();
}
}
}
}
// printNumber.accept(x) outputs "x", where x is an integer.
public void number(IntConsumer printNumber) throws InterruptedException {
synchronized(lock){
while(num<=n){
if(num%3!=0 && num%5!=0){
printNumber.accept(num);
num++;
lock.notifyAll();
}else{
lock.wait();
}
}
}
}
}
解法四:volatile + ReentrantLock (既然synchronized是可以的,那lock必然也是可以的,虽然这两哥几乎也是完全一样的,但是为了复习一下,我们也可以来看看)
class FizzBuzz {
private int n;
private volatile int num =1;
ReentrantLock lock = new ReentrantLock();
Condition condition = lock.newCondition();
public FizzBuzz(int n) {
this.n = n;
}
// printFizz.run() outputs "fizz".
public void fizz(Runnable printFizz) throws InterruptedException {
while(num<=n){
lock.lock();
try{
if(num%3==0 && num%5!=0){
printFizz.run();
num++;
condition.signalAll();
}else{
condition.await();
}
}finally{
lock.unlock();
}
}
}
// printBuzz.run() outputs "buzz".
public void buzz(Runnable printBuzz) throws InterruptedException {
while(num<=n){
lock.lock();
try{
if(num%3!=0 && num%5==0){
printBuzz.run();
num++;
condition.signalAll();
}else{
condition.await();
}
}finally{
lock.unlock();
}
}
}
// printFizzBuzz.run() outputs "fizzbuzz".
public void fizzbuzz(Runnable printFizzBuzz) throws InterruptedException {
while(num<=n){
lock.lock();
try{
if(num%3==0 && num%5==0){
printFizzBuzz.run();
num++;
condition.signalAll();
}else{
condition.await();
}
}finally{
lock.unlock();
}
}
}
// printNumber.accept(x) outputs "x", where x is an integer.
public void number(IntConsumer printNumber) throws InterruptedException {
while(num<=n){
lock.lock();
try{
if(num%3!=0 && num%5!=0){
printNumber.accept(num);
num++;
condition.signalAll();
}else{
condition.await();
}
}finally{
lock.unlock();
}
}
}
}
上面的方法都是使用了传统的互斥来解决问题的,现在让我们来使用一些小东西来直接解决问题,那么都有什么呢
countDownLatch CyclicBarrier semaphore exchanger
解法五:
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();
}
}
}