题目描述：

LeetCode 1115. Print FooBar Alternately

Suppose you are given the following code:

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.

题目大意：

交替输出Foo和Bar。

解题思路：

信号量（Semahpore）

A counting semaphore. Conceptually, a semaphore maintains a set ofpermits. Each acquire blocks if necessary until a permit isavailable, and then takes it. Each release adds a permit,potentially releasing a blocking acquirer.However, no actual permit objects are used; the Semaphore justkeeps a count of the number available and acts accordingly. 

初始化两个信号量semaphoreFoo和semaphoreBar，permits分别为1和0。

foo方法输出Foo之前 acquire semaphoreFoo，之后 release semaphoreBar

bar方法反之。

Java代码：

import java.util.concurrent.Semaphore;
class FooBar {
    private Semaphore semaphoreFoo, semaphoreBar;
    private int n;

    public FooBar(int n) {
        this.semaphoreFoo = new Semaphore(1);
        this.semaphoreBar = new Semaphore(0);
        this.n = n;
    }

    public void foo(Runnable printFoo) throws InterruptedException {

        for (int i = 0; i < n; i++) {
            this.semaphoreFoo.acquire();
            // printFoo.run() outputs "foo". Do not change or remove this line.
            printFoo.run();
            this.semaphoreBar.release();
        }
    }

    public void bar(Runnable printBar) throws InterruptedException {
        
        for (int i = 0; i < n; i++) {
            this.semaphoreBar.acquire();
            // printBar.run() outputs "bar". Do not change or remove this line.
            printBar.run();
            this.semaphoreFoo.release();
        }
    }
}

 

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