题解 | #表示数值的字符串#

class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param str string字符串
     * @return bool布尔型
     */
    bool isNumeric(string str) {

        int index = 0;
        int point_num = 0;
        int num = 0;
        int e_num = 0;
        int E_num = 0;
        while (index < str.size()) {
            //判断有无符号
            //匹配符号
            if (str[index] == '+' || str[index] == '-') {
                if (index + 1 >= str.size() || !((str[index + 1] >= '0' && str[index + 1] <= '9') || str[index + 1] == '.') || (index - 1 >= 0 && str[index - 1] >= '0' && str[index - 1] <= '9')) {
                    return false;
                }
            }
            else if (str[index] == '.') {
                //匹配小数点
                point_num += 1;
                if (point_num > 1)return false;
                if ((index + 1 < str.size() && !(str[index + 1] >= '0' && str[index + 1] <= '9' || str[index + 1] == 'e' || str[index + 1] == 'E')) || (index - 1 >= 0 && !((str[index - 1] >= '0' && str[index - 1] <= '9') || str[index - 1] == '+' || str[index - 1] == '-'))) {
                    return false;
                }
            }
            else if (str[index] == 'e' || str[index] == 'E') {
                //匹配科学计数法
                if(str[index] == 'e')e_num += 1;
                if(str[index] == 'E')E_num += 1;
                if(e_num + E_num > 1)return false;
                if (index - 1 < 0 || !((str[index - 1] >= '0' && str[index - 1] <= '9') || str[index - 1] == '.') ||
                        index + 1 >= str.size() || !((str[index + 1] >= '0' && str[index + 1] <= '9') ||
                                                     (str[index + 1] == '+' || str[index + 1] == '-'))) {
                    return false;
                }
            }
            if(!(str[index] >= '0' && str[index] <= '9' || str[index] == '.' || str[index] == '+' || str[index] == '-' || str[index] == 'e' || str[index] == 'E' || str[index] == ' ')){
            return false;
        }
            if(str[index] >= '0' && str[index] <= '9')num += 1;
            index += 1;
        }

        if(e_num > 0 && point_num > 0 && ((strstr(str.c_str(),".") > strstr(str.c_str(),"e"))))return false;
        if(E_num > 0 && point_num > 0 && ((strstr(str.c_str(),".") > strstr(str.c_str(),"E"))))return false;
        if(num == 0)return false;
        return true;
    }
};

题解：根据规则一一匹配，时间复杂度是O(n)，空间复杂度是O(1)。这样的解法我称之为shi山代码。