# LeetCode Solution, Easy, 1114. Print in Order

# [1114. Print in Order](https://leetcode.com/problems/print-in-order/)

### 題目敘述

Suppose we have a class:

    public class Foo {
      public void first() { print("first"); }
      public void second() { print("second"); }
      public void third() { print("third"); }
    }

The same instance of `Foo` will be passed to three different threads. Thread A will call `first()`, thread B will call `second()`, and thread C will call `third()`. Design a mechanism and modify the program to ensure that `second()` is executed after `first()`, and `third()` is executed after `second()`.

**Note:**

We do not know how the threads will be scheduled in the operating system, even though the numbers in the input seem to imply the ordering. The input format you see is mainly to ensure our tests' comprehensiveness.

**Example 1:**

    Input: [1,2,3]
    Output: "firstsecondthird"
    Explanation: There are three threads being fired asynchronously. The input [1,2,3] means thread A calls first(), thread B calls second(), and thread C calls third(). "firstsecondthird" is the correct output.

**Example 2:**

    Input: [1,3,2]
    Output: "firstsecondthird"
    Explanation: The input [1,3,2] means thread A calls first(), thread B calls third(), and thread C calls second(). "firstsecondthird" is the correct output.

**Constraints:**

- `nums` is a permutation of `[1, 2, 3]`.

#### 題目翻譯

這題很簡單就是需要按照順序去執行 method，按照順序印出字串。

### 解法解析

這題目就是很依賴程式語言的特性，所以可以發現能使用的程式語言相對的少。這邊使用 Python 的標準套件 `threading`，這邊使用的是 `Lock` 來做解題。

#### 程式範例

##### Python

```python
from threading import Lock

class Foo:
    def __init__(self):
        self.firstJobDone = Lock()
        self.secondJobDone = Lock()
        self.firstJobDone.acquire()
        self.secondJobDone.acquire()

    def first(self, printFirst: 'Callable[[], None]') -> None:
        # printFirst() outputs "first".
        printFirst()
        # Notify the thread that is waiting for the first job to be done.
        self.firstJobDone.release()

    def second(self, printSecond: 'Callable[[], None]') -> None:
        # Wait for the first job to be done
        with self.firstJobDone:
            # printSecond() outputs "second".
            printSecond()
            # Notify the thread that is waiting for the second job to be done.
            self.secondJobDone.release()

    def third(self, printThird: 'Callable[[], None]') -> None:

        # Wait for the second job to be done.
        with self.secondJobDone:
            # printThird() outputs "third".
            printThird()
```