1116.Print Zero Even Odd

1116.Print Zero Even Odd

难度:Medium

Suppose you are given the following code:

class ZeroEvenOdd {
  public ZeroEvenOdd(int n) { ... }      // constructor
  public void zero(printNumber) { ... }  // only output 0's
  public void even(printNumber) { ... }  // only output even numbers
  public void odd(printNumber) { ... }   // only output odd numbers
}

The same instance of ZeroEvenOdd will be passed to three different threads:

• Thread A will call zero() which should only output 0's.

• Thread B will call even() which should only ouput even numbers.

• Thread C will call odd() which should only output odd numbers. Each of the thread is given a printNumber method to output an integer. Modify the given program to output the series 010203040506... where the length of the series must be 2n.

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"

也是多锁互斥，这里可以用三把锁，外加一个变量判断输出偶数还是奇数。

class ZeroEvenOdd {
private:
    int n;
    pthread_mutex_t t1= PTHREAD_MUTEX_INITIALIZER;
    pthread_mutex_t t2 = PTHREAD_MUTEX_INITIALIZER;
    pthread_mutex_t t3 = PTHREAD_MUTEX_INITIALIZER;
    bool flag=true;
public:
    ZeroEvenOdd(int n) {
        this->n = n;
        pthread_mutex_lock(&t2) ;
        pthread_mutex_lock(&t3) ;
    }

    // printNumber(x) outputs "x", where x is an integer.
    void zero(function<void(int)> printNumber) {
        for(int i=0;i<n;i++)
        {
            pthread_mutex_lock(&t1);
            printNumber(0);
            flag ?        pthread_mutex_unlock(&t2) :  pthread_mutex_unlock(&t3);

        }

    }

    void even(function<void(int)> printNumber) {
        
        for(int i=1;i<=n;i+=2){
          pthread_mutex_lock(&t2);
          printNumber(i);
          flag=false;
          pthread_mutex_unlock(&t1);
        }
    }

    void odd(function<void(int)> printNumber) {
        for(int i=2;i<=n;i+=2){
        pthread_mutex_lock(&t3);
        flag=true;
        printNumber(i);
        pthread_mutex_unlock(&t1);
        }
    }
};

执行用时 :12 ms, 在所有 C++ 提交中击败了88.99%的用户 内存消耗 :9.1 MB, 在所有 C++ 提交中击败了100.00%的用户