【Leetcode2017.5.18】65. Valid Num

65. Valid Number
    The idea is pretty straightforward. A valid number is composed of the significand and the exponent (which is optional). As we go through the string, do the following things one by one:

skip the leading whitespaces;
check if the significand is valid. To do so, simply skip the leading sign and count the number of digits and the number of points. A valid significand has no more than one point and at least one digit.
check if the exponent part is valid. We do this if the significand is followed by 'e'. Simply skip the leading sign and count the number of digits. A valid exponent contain at least one digit.
skip the trailing whitespaces. We must reach the ending 0 if the string is a valid number.

bool isNumber(const char *s) 
{
    int i = 0;
    
    // skip the whilespaces
    for(; s[i] == ' '; i++) {}
    
    // check the significand
    if(s[i] == '+' || s[i] == '-') i++; // skip the sign if exist
    
    int n_nm, n_pt;
    for(n_nm=0, n_pt=0; (s[i]<='9' && s[i]>='0') || s[i]=='.'; i++)
        s[i] == '.' ? n_pt++:n_nm++;       
    if(n_pt>1 || n_nm<1) // no more than one point, at least one digit
        return false;
    
    // check the exponent if exist
    if(s[i] == 'e') {
        i++;
        if(s[i] == '+' || s[i] == '-') i++; // skip the sign
        
        int n_nm = 0;
        for(; s[i]>='0' && s[i]<='9'; i++, n_nm++) {}
        if(n_nm<1)
            return false;
    }
    
    // skip the trailing whitespaces
    for(; s[i] == ' '; i++) {}
    
    return s[i]==0;  // must reach the ending 0 of the string
}