class FooBar {
public void foo() {
for (int i = 0; i < n; i++) {
print("foo");
}
}
public void bar() {
for (int i = 0; i < n; i++) {
print("bar");
}
}
}
The same instance of FooBar will be passed to two different threads. Thread A will call foo() while thread B will call bar(). Modify the given program to output "foobar" n times.
Example 1:
Input: n = 1
Output: "foobar"
Explanation: There are two threads being fired asynchronously. One of them calls foo(), while the other calls bar(). "foobar" is being output 1 time.
Example 2:
Input: n = 2
Output: "foobarfoobar"
Explanation: "foobar" is being output 2 times.
分析
semaphore(nums)nums设置开始几个线程可以访问,所以foo用的那个要为1
import java.util.concurrent.Semaphore;
class FooBar {
private int n;
Semaphore s1 = new Semaphore(0);;
Semaphore s2 = new Semaphore(1);
public FooBar(int n) {
this.n = n;
}
public void foo(Runnable printFoo) throws InterruptedException {
for (int i = 0; i < n; i++) {
s2.acquire();
// printFoo.run() outputs "foo". Do not change or remove this line.
printFoo.run();
s1.release();
}
}
public void bar(Runnable printBar) throws InterruptedException {
for (int i = 0; i < n; i++) {
s1.acquire();
// printBar.run() outputs "bar". Do not change or remove this line.
printBar.run();
s2.release();
}
}
}