Leetcode 65. Valid Number

https://leetcode.com/problems/valid-number/

One line slick solution: using regular matching.
import re
...
        return bool(re.match("^\s*[\+-]?((\d+(\.\d*)?)|\.\d+)([eE][\+-]?\d+)?\s*$", s))
Run Time is 84 ms.

Another solution is to use states transition method, which is a little hard to get.

Solution:

# T:O(n) S:O(1)
class Solution:
    # @param s, a string
    # @return a boolean
    def isNumber(self, s):
        transition_table = [[-1,  0,  3,  1,  2, -1],     # next states for state 0
                            [-1,  8, -1,  1,  4,  5],     # next states for state 1
                            [-1, -1, -1,  4, -1, -1],     # next states for state 2
                            [-1, -1, -1,  1,  2, -1],     # next states for state 3
                            [-1,  8, -1,  4, -1,  5],     # next states for state 4
                            [-1, -1,  6,  7, -1, -1],     # next states for state 5
                            [-1, -1, -1,  7, -1, -1],     # next states for state 6
                            [-1,  8, -1,  7, -1, -1],     # next states for state 7
                            [-1,  8, -1, -1, -1, -1]]     # next states for state 8
        
        state = 0
        for char in s:
            inputType = InputType.INVALID
            if char.isspace():
                inputType = InputType.SPACE;
            elif char == '+' or char == '-':
                inputType = InputType.SIGN
            elif char.isdigit():
                inputType = InputType.DIGIT
            elif char == '.':
                inputType = InputType.DOT
            elif char == 'e' or char == 'E':
                inputType = InputType.EXPONENT;
                
            state = transition_table[state][inputType];
            
            if state == -1:
                return False;

Run Time: 96 ms