In other words:

• If an oxygen thread arrives at the barrier when no hydrogen threads are present, it has to wait for two hydrogen threads.
• If a hydrogen thread arrives at the barrier when no other threads are present, it has to wait for an oxygen thread and another hydrogen thread.

We don’t have to worry about matching the threads up explicitly; that is, the threads do not necessarily know which other threads they are paired up with. The key is just that threads pass the barrier in complete sets; thus, if we examine the sequence of threads that bond and divide them into groups of three, each group should contain one oxygen and two hydrogen threads.

Write synchronization code for oxygen and hydrogen molecules that enforces these constraints.

Example 1:

Input: "HOH"
Output: "HHO"
Explanation: "HOH" and "OHH" are also valid answers.

Example 2:

Input: "OOHHHH"
Output: "HHOHHO"
Explanation: "HOHHHO", "OHHHHO", "HHOHOH", "HOHHOH", "OHHHOH", "HHOOHH", "HOHOHH" and "OHHOHH" are also valid answers.

Constraints:

• Total length of input string will be 3n, where 1 ≤ n ≤ 30.
• Total number of H will be 2n in the input string.
• Total number of O will be n in the input string.

Solution: Semaphore

class Semaphore {
public:
    Semaphore(int s) : s_(s) {}

    void P(int d = 1) {
      unique_lock<mutex> lock(m_);
      while (s_ < d) {
        cv_.wait(lock);
      }
      s_ -= d;
    }
  
    void V(int d = 1) {      
      unique_lock<mutex> lock(m_);
      s_ += d;      
      cv_.notify_all();
    }
private:
  mutex m_;
  condition_variable cv_;
  int s_;
};

class H2O {
public:
    H2O():s_h_(2), s_o_(2) {  }

    void hydrogen(function<void()> releaseHydrogen) {      
      s_h_.P();
      releaseHydrogen();
      s_o_.V();
    }

    void oxygen(function<void()> releaseOxygen) {
      s_o_.P(2);      
      releaseOxygen();
      s_h_.V(2);
    }
private:
  Semaphore s_h_;
  Semaphore s_o_;
};

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