交替打印FooBar

题目：https://leetcode-cn.com/problems/print-foobar-alternately/submissions/
打印保持顺序，与按序打印类似。

1.用两个信号量解决。

由于c++中没有原生信号量，所以用mutex和条件变量实现一个。

class Semaphore
{
public:
    Semaphore(int value = 1) :count(value) {}

    void Acquire()
    {
        unique_lock<mutex> lck(mtk);
        if (--count < 0)//资源不足挂起线程
            cv.wait(lck);
    }

    void Release()
    {
        unique_lock<mutex> lck(mtk);
        if (++count <= 0)//有线程挂起，唤醒一个
            cv.notify_one();
    }

private:
    int count;
    mutex mtk;
    condition_variable cv;
};

class FooBar {
private:
    int n;
    Semaphore sem1;
    Semaphore sem2;

public:
    FooBar(int n):sem1(1),sem2(0) {
        this->n = n;
    }

    void foo(function<void()> printFoo) {
        for (int i = 0; i < n; ++i) {
            sem1.Acquire();
            // printFoo() outputs "foo". Do not change or remove this line.
            printFoo();
            sem2.Release();
        }
    }

    void bar(function<void()> printBar) {
        for (int i = 0; i < n; ++i) {
            sem2.Acquire();
            // printBar() outputs "bar". Do not change or remove this line.
            printBar();
            sem1.Release();
        }
    }
};

2.用两个互斥锁解决。

二进制信号量的作用与互斥锁功能相似，所以可以直接用mutex代替信号量实现。

class FooBar {
private:
    int n;
    std::mutex foo_mux;
    std::mutex bar_mux;

public:
    FooBar(int n) {
        this->n = n;
        bar_mux.lock();
    }

    void foo(function<void()> printFoo) {
        for (int i = 0; i < n; ++i) {
            foo_mux.lock();
            // printFoo() outputs "foo". Do not change or remove this line.
            printFoo();
            bar_mux.unlock();
        }
    }

    void bar(function<void()> printBar) {
        for (int i = 0; i < n; ++i) {
            bar_mux.lock();
            // printBar() outputs "bar". Do not change or remove this line.
            printBar();
            foo_mux.unlock();
        }
    }
};