65. Valid Number

65. Valid Number (Hard)

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.

Related Topics:
Math, String

Similar Questions:

• String to Integer (atoi) (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/valid-number/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    bool isNumber(string s) {
        int i = 0, N = s.size();
        while (i < N && s[i] == ' ') ++i;
        if (i < N && (s[i] == '+' || s[i] == '-')) ++i;
        bool digitFound = false;
        while (i < N && isdigit(s[i])) { ++i; digitFound = true; }
        if (i < N && s[i] == '.') ++i;
        while (i < N && isdigit(s[i])) { ++i; digitFound = true; }
        if (!digitFound) return false;
        if (i < N && s[i] == 'e') {
            ++i;
            if (i < N && (s[i] == '+' || s[i] == '-')) ++i;
            bool expFound = false;
            while (i < N && isdigit(s[i])) { ++i; expFound = true; }
            if (!expFound) return false;
        }
        while (i < N && s[i] == ' ') ++i;
        return i == N;
    }
};