1115. Print FooBar Alternately

題目敘述

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().

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.

Constraints:

• 1 <= n <= 1000

題目翻譯

題目使用 n 表示要執行多少次，每一次執行都依序執行 foo 和 bar。

解法解析

使用 for-loop 跑 n 次，然後使用 acquire 把程式綁住，等待 release 之後才去執行 print 的操作。

程式範例

Python

from threading import Lock


class FooBar:
    def __init__(self, n):
        self.n = n
        self.foolock = Lock()
        self.barlock = Lock()

    def foo(self, printFoo: 'Callable[[], None]') -> None:
        self.barlock.acquire()
        for i in range(self.n):
            self.foolock.acquire()
            # printFoo() outputs "foo". Do not change or remove this line.
            printFoo()
            self.barlock.release()

    def bar(self, printBar: 'Callable[[], None]') -> None:

        for i in range(self.n):
            self.barlock.acquire()
            # printBar() outputs "bar". Do not change or remove this line.
            printBar()
            self.foolock.release()