Leetcode—1242. 多线程网页爬虫【中等】Plus（多线程）

2024每日刷题（187）

Leetcode—1242. 多线程网页爬虫

实现代码

/**
 * // This is the HtmlParser's API interface.
 * // You should not implement it, or speculate about its implementation
 * class HtmlParser {
 *   public:
 *     vector<string> getUrls(string url);
 * };
 */
class Solution {
public:
    vector<string> crawl(string startUrl, HtmlParser htmlParser) {
        queue<string> q{{startUrl}};
        unordered_set<string> ust{{startUrl}};
        string hostname = getHostName(startUrl);
        vector<thread> threads;
        const int nthreads = std::thread::hardware_concurrency();
        mutex mtx;
        condition_variable cv;

        auto t = [&] {
            while(true) {
                unique_lock<mutex> lock(mtx);
                cv.wait_for(lock, 30ms, [&]() {
                    return q.size();
                });
                if(q.empty()) {
                    return;
                }
                auto cur = q.front();
                q.pop();
                lock.unlock();
                vector<string> urls = htmlParser.getUrls(cur);
                lock.lock();
                for(const string& url: urls) {
                    if(ust.contains(url)) {
                        continue;
                    }
                    if(url.find(hostname) != string::npos) {
                        ust.insert(url);
                        q.push(url);
                    }
                }
                lock.unlock();
                cv.notify_all();
            }
        };

        for(int i = 0; i < nthreads; i++) {
            threads.emplace_back(t);
        }

        for(auto& thread: threads) {
            thread.join();
        }
        return {ust.begin(), ust.end()};
    }
private:
    string getHostName(string& s) {
        int firstIdx = s.find_first_of('/');
        int thirdIdx = s.find_first_of('/', firstIdx + 2);
        return s.substr(firstIdx + 2, thirdIdx - firstIdx - 2);
    }
};

运行结果

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原文地址：https://blog.csdn.net/qq_44631615/article/details/143030798

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