LeetCode 1116. Print Zero Even Odd

原题链接在这里：https://leetcode.com/problems/print-zero-even-odd/

题目：

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

题解：

Use Semaphore. Semaphore(0) means each acquire needs to wait for release.

Semaphore(1) means first acquire could get it, but following acquire needs to wait for release.

Time Complexity: O(1).

Space: O(1).

AC Java:

class ZeroEvenOdd {
    private int n;
    private Semaphore zeroSem;
    private Semaphore evenSem;
    private Semaphore oddSem;
    
    public ZeroEvenOdd(int n) {
        this.n = n;
        zeroSem = new Semaphore(1);
        evenSem = new Semaphore(0);
        oddSem = new Semaphore(0);
    }

    // printNumber.accept(x) outputs "x", where x is an integer.
    public void zero(IntConsumer printNumber) throws InterruptedException {
        for(int i = 1; i <= n; i++){
            zeroSem.acquire();
            printNumber.accept(0);
            if(i % 2 == 0){
                evenSem.release();
            }else{
                oddSem.release();
            }
        }
    }

    public void even(IntConsumer printNumber) throws InterruptedException {
        for(int i = 2; i <= n; i += 2){
            evenSem.acquire();
            printNumber.accept(i);
            zeroSem.release();
        }
    }

    public void odd(IntConsumer printNumber) throws InterruptedException {
        for(int i = 1; i <= n; i += 2){
            oddSem.acquire();
            printNumber.accept(i);
            zeroSem.release();
        }
    }
}

类似Fizz Buzz Multithreaded.