65. Valid Number

• Difficulty: Hard

• Topics: Math, String

• Similar Questions:

  • String to Integer (atoi)

Problem:

Validate if a given string can be interpreted as a decimal number.

Some examples:
"0" => true
" 0.1 " => true
"abc" => false
"1 a" => false
"2e10" => true
" -90e3   " => true
" 1e" => false
"e3" => false
" 6e-1" => true
" 99e2.5 " => false
"53.5e93" => true
" --6 " => false
"-+3" => false
"95a54e53" => false

Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. However, here is a list of characters that can be in a valid decimal number:

• Numbers 0-9
• Exponent - "e"
• Positive/negative sign - "+"/"-"
• Decimal point - "."

Of course, the context of these characters also matters in the input.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.

Solutions:

class Solution {
public:
    bool isNumber(string s) {
        bool sign = false;
        bool dot = false;
        bool exponent = false;
        bool num = false;

        int left = 0;
        while (left < s.length() && s[left] == ' ')  ++left;

        int right = s.length() - 1;
        while (right >= 0 && s[right] == ' ') --right;

        if (left > right)   return false;

        for (int i = left; i <= right; ++i) {
            char c = s[i];
            if (isdigit(c)) {
                sign = true;
                num = true;
               // if (c == '0' && i -1 >= 0 && s[i-1] == '0' && !dot) return false;
            }            
            else if (c == '+' || c == '-') {
                if (sign)   return false;
                sign = true;
            }            
            else if (c == 'e') {
                if (exponent)   return false;
                if (!num)   return false;
                sign = false;
                dot = true;
                exponent = true;
                num = false;
            }  
            else if (c == '.') {
                if (dot)    return false;
                dot = true;
                sign = true;
            }
            else {
                return false;
            }
        }

        return s[right] != 'e' && num;
    }
};