# 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`

### 解法解析

#### 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()
``````

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