Leetcode:65. Valid Number

Description

Validate if a given string is numeric.

Some examples:
"0" => true
" 0.1 " => true
"abc" => false
"1 a" => false
"2e10" => true
Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one.

思路

• 这个题简直就是恶心。按照剑指offer上的思路来做，简直不要太坑。

代码

class Solution {
public:
    bool isNumber(string s) {
        int len = s.size();
        if(len == 0) return false;
        
        int i = 0;
        while(i < len && s[i] == ' ') i++;
        if(i < len && (s[i] == '+' || s[i] == '-'))
            i++;
        if(i == len) return false;
        
        bool digit = false, digit2 = false;
        if(s[i] >= '0' && s[i] <= '9'){
            digit = true;
            while(i < len && s[i] >= '0' && s[i] <= '9') i++;
        }
        
        bool isExpon = false, judge = false, isD = false;
        if(i < len && s[i] == '.'){
            isD = true;
            i++;
            while(i < len && s[i] >= '0' && s[i] <= '9'){
                digit2 = true;
                i++;
            }
            
            
            if(i < len && (s[i] == 'e' || s[i] == 'E')){
                judge = true;
                isExpon = isValidOfE(s, i, len);
            }
            
            while(i < len && s[i] == ' ')
                i++;
        }
        else if(i < len && (s[i] == 'e' || s[i] == 'E')){
            judge = true;
            isExpon = isValidOfE(s, i, len);
        }
        
        while(i < len && s[i] == ' ')
            i++;
            
        if(i != len)
            return false;
            
         if(isD && !digit && !digit2)
            return false;
     
        if(judge){
            if(!isExpon) return false;
            
            return (digit || (isD && digit2));
        }
        return true;
    }
    
    bool isValidOfE(string& str, int &i, int len){
        if(str[i] == 'e' || str[i] == 'E')
            i++;
        
        if(i < len && (str[i] == '-' || str[i] == '+'))
            i++;
        
        if(i == len) return false;
        
        if(str[i] >= '0' && str[i] <= '9'){
            while(i < len && (str[i] >= '0' && str[i] <= '9'))
                i++;
                
            return true;
        }
        else return false;
    }
};