acwing 31等价于LeetCode 65 有效数字
功能上类似于leetcode 8

/*
思路1：模拟法
    方法1：两段法，以e/E为分界点
    方法2：三段法，以.和e/E为分界点
思路2：正则表达式
思路3：有限确定状态机DFA
正则文法、正则表达式、DFA可以相互转化。
https://leetcode-cn.com/problems/valid-number/solution/biao-qu-dong-fa-by-user8973/
*/
/*思路1-1 模拟法*/
class Solution {
public:
    bool isNumber(string s) {
        size_t e_pos = s.find('e');
        if (e_pos == string::npos)
            e_pos = s.find('E');
        if (e_pos == string::npos)
            return isPureNumber(s.substr(0, e_pos), false);

        return isPureNumber(s.substr(0, e_pos), false)
            && isPureNumber(s.substr(e_pos + 1), true);
    }
    bool isPureNumber(const string &s, bool no_dot) {
        // 返回值代表是否包含数字
        // no_dot为ture，s中不能含有dot
        if (s.empty()) return false;

        int st = int(s[0]=='+' || s[0]=='-');
        bool has_digit = false;
        for (int i = st; i < s.size(); i ++) {
            if (s[i]=='.') {
                if (no_dot) return false;
                no_dot = true;
            }
            else if (isdigit(s[i])) has_digit = true;
            else return false;
        }
        return has_digit;
    }
};
/*思路1-2 模拟法*/
class Solution {
public:
    bool isNumber(string s) {
        int i = 0;
        while (i < s.size() && s[i] == ' ') i ++ ;
        bool num1 = false, num2 = false, num3 = true;

        num1 = scanInteger(s, i);
        if (s[i] == '.') num2 = scanUnsignedInteger(s,  ++ i);
        if (s[i] == 'e' || s[i] == 'E') num3 = scanInteger(s,  ++ i);

        while (i < s.size() && s[i] == ' ') i ++ ;
        // printf("%d %d %d",num1, num2, num3);
        return (num1 || num2) && num3 && i == s.size();
    }

    bool scanInteger(string& s, int& pos) {
        if (s[pos] == '+' || s[pos] == '-') pos ++ ;
        return scanUnsignedInteger(s, pos);
    }

    bool scanUnsignedInteger(string& s, int& pos) {
        int p = pos;
        while (pos < s.size() && s[pos] >= '0' && s[pos] <= '9') pos ++ ;
        return pos > p;
    }
};
/*思路2 正则表达式*/
class Solution {
public:
    bool isNumber(string s) {
        string pattern = "(\\+|\\-)?((\\d+\\.?)|(\\.?\\d+)|(\\d+\\.\\d+))((e|E)(\\+|\\-)?\\d+)?";
        string pattern2 = "(\\+|\\-)?((\\d+)|(\\d+\\.)|(\\.\\d+)|(\\d+\\.\\d+))((e|E)(\\+|\\-)?\\d+)?";
        return regex_match(s, regex(pattern));
    }
};
/*思路3 DFA*/
class Solution {
public:
    bool isNumber(string s) {
        if(s.empty()) return false;
        int n = s.size();

        int state = 0;
        vector<bool> finals({0, 0, 0, 1, 0, 1, 1, 0, 1}); // 合法的终止状态
        vector<vector<int> > transfer({
            {0,  1,  6,  2,  -1, -1},
            {-1, -1, 6,  2,  -1, -1},
            {-1, -1, 3,  -1, -1, -1},
            {8,  -1, 3,  -1, 4,  -1},
            {-1, 7,  5,  -1, -1, -1},
            {8,  -1, 5,  -1, -1, -1},
            {8,  -1, 6,  3,  4,  -1},
            {-1, -1, 5,  -1, -1, -1},
            {8,  -1, -1, -1, -1, -1},
        });

        for(int i = 0; i < n; ++i) {
            state = transfer[state][_make(s[i])];
            if(state < 0) return false;
        }
        return finals[state];
    }

private:
    int _make(const char& c) {
        switch(c) {
            case ' ': return 0;
            case '+': return 1;
            case '-': return 1;
            case '.': return 3;
            case 'e': return 4;
            case 'E': return 4;
            default: return _number(c);
        }
    }

    int _number(const char& c) {
        if(c >= '0' && c <= '9') return 2;
        else return 5;
    }
};