# LeetCode Solution, Medium, 1115. Print FooBar Alternately

# [1115. Print FooBar Alternately](https://leetcode.com/problems/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

```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()
```