题目

状态机

很容易有一种思路就是状态机，因为这题状态很少，只有两个，实体，非实体，编译器和解释器本身就是个状态机

cpp
class Solution {
public:
    string entityParser(string text) {
        int state = 0;
        string result = "";
        string entity = "";

        for (char ch : text) {
            switch (state) {
                case 0:
                    if (ch == '&') {
                        state = 1;
                        entity += ch;
                    } else {
                        result += ch;
                    }
                    break;
                case 1:
                    if (ch == '&') {
                        result += entity; // 不是实体，加入结果
                        entity = "&"; // 重置entity
                    } else if (ch == ';') {
                        entity += ch;
                        result += replaceEntity(entity); // 替换实体并加入结果
                        entity = ""; // 重置entity
                        state = 0;
                    } else {
                        entity += ch;
                    }
                    break;
            }
        }

        if (state == 1) {
            result += entity; // 处理结尾的未结束实体
        }

        return result;
    }

private:
    string replaceEntity(string entity) {
        if (entity == """) return "\"";
        if (entity == "'") return "'";
        if (entity == "&") return "&";
        if (entity == ">") return ">";
        if (entity == "&lt;") return "<";
        if (entity == "&frasl;") return "/";
        return entity; // 未知实体，保持不变
    }
};

字符串替换

cpp
class Solution {
public:
 string entityParser(string text) {
        unordered_map<string, string> entityMap = {
            {"&quot;", "\""},
            {"&apos;", "'"},
            {"&amp;", "&"},
            {"&gt;", ">"},
            {"&lt;", "<"},
            {"&frasl;", "/"}
        };

        for (auto& entry : entityMap) {
            size_t pos = 0;
            while ((pos = text.find(entry.first, pos)) != string::npos) {
                text.replace(pos, entry.first.length(), entry.second);
                pos += entry.second.length();
            }
        }

        return text;
    }
};