最短摘要与反转链表

关于最短摘要，编程之美3.5节并没有给出详细的代码。其实还是值得一写的。下面的实现我使用了set和map来维护相关的信息。

int main() {
    int n, m;
    while (cin>>n>>m) {
        // input
        vector<string> seq;
        while (n--) *back_inserter(seq) = *istream_iterator<string>(cin);
        set<string> kwords;
        while (m--) *inserter(kwords, kwords.end()) = *istream_iterator<string>(cin);

        // find shortest abstract
        typedef vector<string>::iterator Vsit;
        pair<Vsit, Vsit> r(seq.begin(), seq.begin()), q = r;

        set<string> notfound = kwords;
        map<string, int> found;
        for(;;) {
            if (notfound.size()) {
                if (q.second == seq.end()) break;

                set<string>::iterator it = notfound.find(*q.second);
                if (it != notfound.end()) {
                    ++found[*it];
                    notfound.erase(it);
                }
                else {
                    map<string, int>::iterator it2 = found.find(*q.second);
                    if (it2 != found.end()) ++(it2->second);
                }

                ++(q.second);
            }
            else {
                map<string, int>::iterator it = found.find(*q.first);
                if (it != found.end() && !--(it->second)) {
                    size_t rlen = distance(r.first, r.second);
                    size_t qlen = distance(q.first, q.second);
                    if (!rlen || rlen > qlen) r = q;

                    notfound.insert(it->first);
                    found.erase(it);
                }

                ++(q.first);
            }
        }

        // output
        if (!distance(r.first, r.second)) cout<<"No abstract available.";
        else {
            if (r.first != seq.begin()) cout<<"... ";
            for (bool first = true; r.first != r.second; ++r.first, first = false) {
                if (!first) cout<<' ';
                cout<<*r.first;
            }
            if (r.second != seq.end()) cout<<" ...";
        }
        cout<<'\n';
    }
    return 0;
}

顺便，把3.4节中的扩展问题单向链表反转实现也列在下面：

Node* reverse(Node* list) {
    Node* prev = 0;
    while (list) {
        Node* t = list;
        list = list->next;
        t->next = prev;
        prev = t;
    }
    return prev;
}