Leetcode c++ 1114. 按序打印 1114. Print in Order

本文探讨了C++中实现线程同步的多种方法,包括使用计数变量、future和promise、互斥量、unique_lock和lock_guard等,以确保线程按特定顺序执行。

1114. 按序打印
我们提供了一个类:

public class Foo {
public void one() { print(“one”); }
public void two() { print(“two”); }
public void three() { print(“three”); }
}
三个不同的线程将会共用一个 Foo 实例。

线程 A 将会调用 one() 方法
线程 B 将会调用 two() 方法
线程 C 将会调用 three() 方法
请设计修改程序,以确保 two() 方法在 one() 方法之后被执行,three() 方法在 two() 方法之后被执行。

示例 1:

输入: [1,2,3]
输出: “onetwothree”
解释:
有三个线程会被异步启动。
输入 [1,2,3] 表示线程 A 将会调用 one() 方法,线程 B 将会调用 two() 方法,线程 C 将会调用 three() 方法。
正确的输出是 “onetwothree”。
示例 2:

输入: [1,3,2]
输出: “onetwothree”
解释:
输入 [1,3,2] 表示线程 A 将会调用 one() 方法,线程 B 将会调用 three() 方法,线程 C 将会调用 two() 方法。
正确的输出是 “onetwothree”。

没有接触过多线程编程的我写的代码,提交了两次才通过,第一次甚至超时:

class Foo {
public:
    Foo() {
        
    }
    int count=0;
    void first(function<void()> printFirst) {
        // printFirst() outputs "first". Do not change or remove this line.
        printFirst();
        count=1;
    }

    void second(function<void()> printSecond) {
        while(count<=0){
            
        }
        // printSecond() outputs "second". Do not change or remove this line.
        printSecond();
        count=2;
    }

    void third(function<void()> printThird) {
        while(count<=1){
            
        }
        // printThird() outputs "third". Do not change or remove this line.
        printThird();
    }
};

其他方法,一些新东西

  1. future 头文件和promise
#include <future>

class Foo {
public:
    promise<void> p1, p2;
    
    Foo() {
        
    }

    void first(function<void()> printFirst) {
        
        // printFirst() outputs "first". Do not change or remove this line.
        printFirst();
        p1.set_value(); // 为p1赋值
    }

    void second(function<void()> printSecond) {
        p1.get_future().wait(); // 等待p1
        // printSecond() outputs "second". Do not change or remove this line.
        printSecond();
        p2.set_value(); // 为p2赋值
    }

    void third(function<void()> printThird) {
        p2.get_future().wait(); // 等待p2
        // printThird() outputs "third". Do not change or remove this line.
        printThird();
    }
};

  1. C++互斥量pthread_mutex_t
class Foo {
public:
    pthread_mutex_t one;
    pthread_mutex_t two;
    Foo() {
        pthread_mutex_init(&one, NULL);
        pthread_mutex_init(&two, NULL);
        pthread_mutex_lock(&one);
        pthread_mutex_lock(&two);
    }

    void first(function<void()> printFirst) {
        
        // printFirst() outputs "first". Do not change or remove this line.
        printFirst();
        pthread_mutex_unlock(&one);
    }

    void second(function<void()> printSecond) {
        
        // printSecond() outputs "second". Do not change or remove this line.
        pthread_mutex_lock(&one);
        printSecond();
        pthread_mutex_unlock(&one);
        pthread_mutex_unlock(&two);
    }

    void third(function<void()> printThird) {
        
        // printThird() outputs "third". Do not change or remove this line.
        pthread_mutex_lock(&two);
        printThird();
        pthread_mutex_unlock(&two);
    }
};

3.C++ 并发编程,std::unique_lock与std::lock_guard

class Foo {
public:
    Foo() {
        
    }

    void first(function<void()> printFirst) {
        // 等待直至 main() 发送数据
	    std::unique_lock<std::mutex> lk(g_mutex);  
        // printFirst() outputs "first". Do not change or remove this line.
        printFirst();
        // 通知前完成手动解锁,以避免等待线程才被唤醒就阻塞(细节见 notify_one )
        counter++;
        cv1.notify_one();
    }

    void second(function<void()> printSecond) {
	    std::unique_lock<std::mutex> lk(g_mutex);
        cv1.wait(lk, [this](){return counter == 2;}); // 阻塞当前线程,直到条件变量被唤醒 
        // printSecond() outputs "second". Do not change or remove this line.
        printSecond();
        counter++;
        cv2.notify_one();
    }

    void third(function<void()> printThird) {
        
	    std::unique_lock<std::mutex> lk(g_mutex);
        cv2.wait(lk, [this](){return counter == 3;});
        // printThird() outputs "third". Do not change or remove this line.
        printThird();
    }
    
private:
    int counter = 1;
    std::condition_variable cv1;
    std::condition_variable cv2;
    // 使用lock和unlock手动加锁
    std::mutex g_mutex;
};

4.还有mutex 等,互斥变量。。还蛮多的

B. Array Recoloring time limit per test2 seconds memory limit per test256 megabytes You are given an integer array a of size n . Initially, all elements of the array are colored red. You have to choose exactly k elements of the array and paint them blue. Then, while there is at least one red element, you have to select any red element with a blue neighbor and make it blue. The cost of painting the array is defined as the sum of the first k chosen elements and the last painted element. Your task is to calculate the maximum possible cost of painting for the given array. Input The first line contains a single integer t (1≤t≤103 ) — the number of test cases. The first line of each test case contains two integers n and k (2≤n≤5000 ; 1≤k<n ). The second line contains n integers a1,a2,…,an (1≤ai≤109 ). Additional constraint on the input: the sum of n over all test cases doesn't exceed 5000 . Output For each test case, print a single integer — the maximum possible cost of painting for the given array. Example InputCopy 3 3 1 1 2 3 5 2 4 2 3 1 3 4 3 2 2 2 2 OutputCopy 5 10 8 Note In the first example, you can initially color the 2 -nd element, and then color the elements in the order 1,3 . Then the cost of painting is equal to 2+3=5 . In the second example, you can initially color the elements 1 and 5 , and then color the elements in the order 2,4,3 . Then the cost of painting is equal to 4+3+3=10 . In the third example, you can initially color the elements 2,3,4 , and then color the 1 -st element. Then the cost of painting is equal to 2+2+2+2=8 . 用cpp解决
最新发布
03-19
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