LeetCode: 1116. 打印零与奇偶数
Semaphore 信号量
通过信号量模拟,控制管理资源
Semaphore s = new Semaphore(0);
Semaphore 内部维护了一个计数值
计数值为0,表示线程进入休眠,一开始就阻塞,让其他线程完成执行。
acquire() 方法的使用会使线程阻塞,直到获取到新的许可
release() 方法释放所有持有许可的线程,并返回一个新的可用许可给 Semaphore
参考博客:
public class Thread1116 {
private int n;
// 信号量模拟
// 构造器 --- 许可管理资源的数量
private Semaphore zero = new Semaphore(1);
private Semaphore even = new Semaphore(0); // 一开始使线程阻塞,从而完成其他的线程执行
private Semaphore odd = new Semaphore(0); // 同上
public Thread1116(int n) {
this.n = n;
}
// printNumber.accept(x) outputs "x", where x is an integer.
public void zero(IntConsumer printNumber) throws InterruptedException {
for (int i = 1; i <= n; i++) {
// 有几个数,输出几个 0
// 同时需要控制 奇数、偶数的输出线程切换
zero.acquire(); // 阻塞线程
printNumber.accept(0);
// 奇数
if(i % 2 == 1){
odd.release();
}else {
even.release();
}
}
}
public void even(IntConsumer printNumber) throws InterruptedException {
for (int i = 2; i <= n; i += 2) {
even.acquire();
printNumber.accept(i);
zero.release();
}
}
public void odd(IntConsumer printNumber) throws InterruptedException {
for (int i = 1; i <= n; i += 2) {
odd.acquire();
printNumber.accept(i);
zero.release();
}
}
public static void main(String[] args) {
Thread1116 t = new Thread1116(5);
new Thread(new Runnable() {
@Override
public void run() {
try {
t.zero(new IntConsumer() {
@Override
public void accept(int value) {
System.out.println(value);
}
});
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}).start();
new Thread(new Runnable() {
@Override
public void run() {
try {
t.even(new IntConsumer() {
@Override
public void accept(int value) {
System.out.println(value);
}
});
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}).start();
new Thread(new Runnable() {
@Override
public void run() {
try {
t.odd(new IntConsumer() {
@Override
public void accept(int value) {
System.out.println(value);
}
});
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}).start();
}
}