通配符、正则表达式的回溯解法

最新推荐文章于 2025-02-04 09:29:59 发布

杨葳

最新推荐文章于 2025-02-04 09:29:59 发布

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分类专栏： C++ 文章标签： string 正则表达式 recursion

本文链接：https://blog.youkuaiyun.com/qq_25310869/article/details/68936659

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bool recursion_util(string& text, string& pattern, int pos1, int pos2)
{
    if (pos2 == pattern.size())
        return pos1 == text.size();
    if (pattern.at(pos2) != '*') {
        if (pos1 < text.size() && (text.at(pos1) == pattern.at(pos2) || pattern.at(pos2) == '?'))
            return recursion_util(text, pattern, pos1+1, pos2+1);
        else
            return false;
    }
    else {
        if ((pos2+1) < text.size() && pattern.at(pos2+1) == '*') {
            pos2++;
        }
        while (pos1 < text.size())
        {
            if (recursion_util(text, pattern, pos1, pos2+1))
                return true;
            pos1++;
        }
        return false;
    }
}
bool recursion_wild_card(string text, string pattern) {
    return recursion_util(text, pattern, 0, 0);
}



bool recursion_regular_util(string& text, string& pattern, int pos1, int pos2)
{
    if (pos2 == pattern.size())
        return pos1 == text.size();
    if (pattern.at(pos2) != '*') {
        if (pos1 < text.size() && (text.at(pos1) == pattern.at(pos2) || pattern.at(pos2) == '.'))
            return recursion_regular_util(text, pattern, pos1+1, pos2+1);
        return false;
    }
    else {
        while (pos2 < pattern.size()-1 && pattern.at(pos2+1) == '*')
            pos2++;

        pos1--;
        while (pos1 < text.size())
        {
            if (recursion_regular_util(text, pattern, pos1, pos2+1))
                return text.at(pos1) == pattern.at(pos2-1) || pattern.at(pos2-1) == '.';
            pos1++;
        }
        return false;
    }
}

bool recursion_regular(string text, string pattern) 
{
    return recursion_regular_util(text, pattern, 0, 0);
}