Leetcode 1115 交替打印 FooBar ( Print FooBar Alternately *Medium* ) 题解分析

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无聊随机了道并发题，还是比较明显的用信号量就可以解决的题，不过第一次做还是有点不适应，要加油多练练

无聊想去 roll 一题就看到了有并发题，就找到了这题，其实一眼看我的想法也是用信号量，但是用 condition 应该也是可以处理的，不过这类问题好像本地有点难调，因为它好像是抽取代码执行的，跟直观的逻辑比较不一样
Suppose you are given the following code:

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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.

示例

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.

题解

简析

其实用信号量是很直观的，就是让打印 foo 的线程先拥有信号量，打印后就等待，给 bar 信号量 + 1，然后 bar 线程运行打印消耗 bar 信号量，再给 foo 信号量 + 1

code

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class FooBar {
    
    private final Semaphore foo = new Semaphore(1);
    private final Semaphore bar = new Semaphore(0);
    private int n;

    public FooBar(int n) {
        this.n = n;
    }

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

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