65. Valid Number

class Solution {
public:
    bool isNumber(string s) {
        int len = s.length();
        if(len == 0) return false;
        int start = 0;
        while(s[start] == 32) //space
        {
            start++; //neglect space at the beginning
        }
        if(start >= len) return false; //false if only congtaining space
        
        int end = len-1;
        while(s[end] == 32)
        {
            end--;//neglect space at the end
        }
        bool hasDot = false;
        bool hasE = false;
        bool hasDigit1 = false; //if digit appear before e
        bool hasDigit2 = false; //if digit appear after e
        for(int i = start; i<=end; i++)
        {
            if(s[i] ==43 || s[i] == 45) //+, -
            {
                if(i!=start && s[i-1]!= 101) return false;//false if +,- doesn‘t appear in the beginning, or behind e
            }
            else if(s[i] == 46) //. 
            {
                if(hasDot) return false; //false if two . appear
                if(hasE) return false; //. cannot appear behind e
                hasDot = true;
            }
            else if(s[i]== 101)  //e
            {
                if(hasE) return false; //false if appears two e
                if(!hasDigit1) return false; //false if there‘s no digit before e
                hasE = true;
            }
            else if(s[i]< 48 || s[i] > 57) return false;
            else if(!hasDigit1) hasDigit1 = true;
            else if(hasE) hasDigit2 = true;
        }
        if((hasE && hasDigit1 && hasDigit2) || (!hasE && hasDigit1)) return true; //at lease one digit must exist; if e appear, at least one digit must appear both before and after e
        else return false;
    }
};