# LeetCode Solution, Medium, 1116. Print Zero Even Odd

# [1116. Print Zero Even Odd](https://leetcode.com/problems/print-zero-even-odd/)

### 題目敘述

You have a function `printNumber` that can be called with an integer parameter and prints it to the console.

- For example, calling `printNumber(7)` prints `7` to the console.

You are given an instance of the class `ZeroEvenOdd` that has three functions: `zero`, `even`, and `odd`. The same instance of `ZeroEvenOdd` will be passed to three different threads:

- **Thread A**: calls `zero()` that should only output `0`'s.
- **Thread B**: calls `even()` that should only output even numbers.
- **Thread C**: calls `odd()` that should only output odd numbers.

Modify the given class to output the series `"010203040506..."` where the length of the series must be `2n`.

Implement the `ZeroEvenOdd` class:

- `ZeroEvenOdd(int n)` Initializes the object with the number `n` that represents the numbers that should be printed.
- `void zero(printNumber)` Calls `printNumber` to output one zero.
- `void even(printNumber)` Calls `printNumber` to output one even number.
- `void odd(printNumber)` Calls `printNumber` to output one odd number.

**Example 1:**

    Input: n = 2
    Output: "0102"
    Explanation: There are three threads being fired asynchronously. One of them calls zero(), the other calls even(), and the last one calls odd(). "0102" is the correct output.

**Example 2:**

    Input: n = 5
    Output: "0102030405"

**Constraints:**

- `1 <= n <= 1000`

#### 題目翻譯

這題的條件是，會有三個 method，分別是 zero, even, odd。會給一個參數 `n`，然後需要依序從 1 印到 `n`，但是在每個數字前都還要先印出 0。如果是印出 0 則要執行 `zero`，印出偶數需要使用 `even`，奇數則要使用 `odd`。

### 解法解析

類似上一題 1115 一樣，使用 for-loop 去處理不同 method 的執行時機。每次 `even` 或 `odd` 執行完後，都回到 `zero` 的狀態。然後在 `zero` 的 method 中去判斷下次要執行的是 `odd` 或 `even`。

#### 程式範例

##### Python

```python
from threading import Lock


class ZeroEvenOdd:
    def __init__(self, n):
        self.n = n

        self.zero_mutex = Lock()
        self.odd_mutex = Lock()
        self.even_mutex = Lock()

        self.even_mutex.acquire()
        self.odd_mutex.acquire()

    # printNumber(x) outputs "x", where x is an integer.
    def zero(self, printNumber: 'Callable[[int], None]') -> None:
        for i in range(self.n):
            self.zero_mutex.acquire()

            printNumber(0)

            if (i % 2) == 0:
                self.even_mutex.release()
            else:
                self.odd_mutex.release()

    def even(self, printNumber: 'Callable[[int], None]') -> None:
        for i in range(2, self.n + 1, 2):
            self.odd_mutex.acquire()
            printNumber(i)
            self.zero_mutex.release()
        self.odd_mutex.release()

    def odd(self, printNumber: 'Callable[[int], None]') -> None:
        for i in range(1, self.n + 1, 2):
            self.even_mutex.acquire()
            printNumber(i)
            self.zero_mutex.release()
        self.even_mutex.release()
```