难度:Medium
Suppose you are given the following code:
class FooBar { public void foo() { for (int i = 0; i < n; i++) { print("foo"); } }
public void bar() { for (int i = 0; i < n; i++) { print("bar"); } } } The same instance of FooBar will be passed to two different threads. Thread A will call foo() while thread B will call bar(). Modify the given program to output "foobar" n times.
Example 1:Input: n = 1Output: "foobar"Explanation: There are two threads being fired asynchronously. One of them calls foo(), while the other calls bar(). "foobar" is being output 1 time.Example 2:Input: n = 2Output: "foobarfoobar"Explanation: "foobar" is being output 2 times.
要交替输出foo和bar,在每个函数中,每次循环结束将当前线程锁起来,令下一个循环处于阻塞状态,待另一个函数的当前循环结束后,解锁,交替进行输出。
class FooBar {private:int n;pthread_mutex_t t1=PTHREAD_MUTEX_INITIALIZER;pthread_mutex_t t2=PTHREAD_MUTEX_INITIALIZER;public:FooBar(int n) {this->n = n;pthread_mutex_lock(&t1);}void foo(function<void()> printFoo) {for (int i = 0; i < n; i++) {pthread_mutex_lock(&t2);// printFoo() outputs "foo". Do not change or remove this line.printFoo();pthread_mutex_unlock(&t1);}}void bar(function<void()> printBar) {for (int i = 0; i < n; i++) {pthread_mutex_lock(&t1);// printBar() outputs "bar". Do not change or remove this line.printBar();pthread_mutex_unlock(&t2);}}};
执行用时 :52 ms, 在所有 C++ 提交中击败了50.00%的用户 内存消耗 :10.3 MB, 在所有 C++ 提交中击败了100.00%的用户