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.
直接上代码，特别是解法二，引入 re 模块匹配，很精简强大！
1 class InputType: 2 INVALID = 0 3 SPACE = 1 4 SIGN =2 5 DIGIT =3 6 DOT = 4 7 EXPONENT =5 8 9 class Solution: 10 # @param s, a string 11 # @return a boolean 12 def isNumber(self, s): 13 transition_table = [[-1, 0, 3, 1, 2, -1], # next states for state 0 14 [-1, 8, -1, 1, 4, 5], # next states for state 1 15 [-1, -1, -1, 4, -1, -1], # next states for state 2 16 [-1, -1, -1, 1, 2, -1], # next states for state 3 17 [-1, 8, -1, 4, -1, 5], # next states for state 4 18 [-1, -1, 6, 7, -1, -1], # next states for state 5 19 [-1, -1, -1, 7, -1, -1], # next states for state 6 20 [-1, 8, -1, 7, -1, -1], # next states for state 7 21 [-1, 8, -1, -1, -1, -1]] # next states for state 8 22 23 state = 0 24 for char in s: 25 inputType = InputType.INVALID 26 if char.isspace(): 27 inputType = InputType.SPACE; 28 elif char == '+' or char == '-': 29 inputType = InputType.SIGN 30 elif char in '0123456789': 31 inputType =InputType.DIGIT 32 elif char == '.': 33 inputType = InputType.DOT 34 elif char == 'e' or char =='E': 35 inputType = InputType.EXPONENT 36 37 state = transition_table[state][inputType] 38 if state == -1: 39 return False 40 return state==1 or state==4 or state ==7 or state==8 41 42 def isNumber2(self,s): 43 import re 44 return bool(re.match("^s*[+-]?((d+(.d*)?)|.d+)([eE][+-]?d+)?s*$",s))