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1116.Print Zero Even Odd

1116.Print Zero Even Odd

难度:Medium
Suppose you are given the following code:
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class ZeroEvenOdd {
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public ZeroEvenOdd(int n) { ... } // constructor
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public void zero(printNumber) { ... } // only output 0's
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public void even(printNumber) { ... } // only output even numbers
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public void odd(printNumber) { ... } // only output odd numbers
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}
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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.
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Example 1:
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Input: n = 2
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Output: "0102"
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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.
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Example 2:
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Input: n = 5
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Output: "0102030405"
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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。
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class ZeroEvenOdd {
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private:
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int n;
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pthread_mutex_t t1= PTHREAD_MUTEX_INITIALIZER;
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pthread_mutex_t t2 = PTHREAD_MUTEX_INITIALIZER;
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pthread_mutex_t t3 = PTHREAD_MUTEX_INITIALIZER;
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bool flag=true;
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public:
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ZeroEvenOdd(int n) {
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this->n = n;
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pthread_mutex_lock(&t2) ;
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pthread_mutex_lock(&t3) ;
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}
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// printNumber(x) outputs "x", where x is an integer.
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void zero(function<void(int)> printNumber) {
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for(int i=0;i<n;i++)
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{
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pthread_mutex_lock(&t1);
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printNumber(0);
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flag ? pthread_mutex_unlock(&t2) : pthread_mutex_unlock(&t3);
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}
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}
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void even(function<void(int)> printNumber) {
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for(int i=1;i<=n;i+=2){
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pthread_mutex_lock(&t2);
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printNumber(i);
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flag=false;
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pthread_mutex_unlock(&t1);
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}
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}
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void odd(function<void(int)> printNumber) {
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for(int i=2;i<=n;i+=2){
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pthread_mutex_lock(&t3);
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flag=true;
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printNumber(i);
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pthread_mutex_unlock(&t1);
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}
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}
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};
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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 2yr ago
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