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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
Last modified 1yr ago