C++ 代码

class Solution {
public:
    bool isNumber(string s) {
        int countE = 0;
        int countP = 0;
        int countO = 0;
        int count = 0;
        int indexE[100];
        int indexP[100];
        int indexO[100];
        if (s[0] == 'e' || s[0] == 'E') return 0;
        if (s[0] == '.' && (s[1] == 'e' || s[1] == 'E') ) return 0;
        if (s[s.size()-1] == 'e') return 0;
        for (int i = 0; i < s.size(); i++)
        {  
            if ( !( ( s[i] <= '9' && s[i] >= '0') || s[i] == '+' || s[i] == '-' || s[i] == 'E' || s[i] == 'e' || s[i] == '.' ))return 0;
            if ( s[i] == '+' || s[i] == '-') indexO[countO++] = i;
            if ( s[i] == 'E' || s[i] == 'e') indexE[countE++] = i;
            if ( s[i] == '.') indexP[countP++] = i;
            if ( (s[i] == '+' || s[i] =='-' ) && (s[i+1] == '+' || s[i+1] =='-')) return 0;
            if ( countP > 1 || countE > 1) return 0;
            count++;
        }
        if ( count - countE - countP - countO ==0) return 0;
        if ( countO == count || countE == count || countP == count) return 0 ;
        if ( countO && countP && countE && (indexP[countP - 1] > indexE[countE - 1]) ) return 0;
        if ( !countE && countO && indexO[0] != 0) return 0;
        return 1;
    }
};