难度:Medium
Suppose you are given the following code:
class ZeroEvenOdd {public ZeroEvenOdd(int n) { ... } // constructorpublic void zero(printNumber) { ... } // only output 0'spublic void even(printNumber) { ... } // only output even numberspublic 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 = 2Output: "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 = 5Output: "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