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valid-number.py
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valid-number.py
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# Time: O(n)
# Space: O(1)
#
# 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.
#
class InputType:
INVALID = 0
SPACE = 1
SIGN = 2
DIGIT = 3
DOT = 4
EXPONENT = 5
# regular expression: "^\s*[\+-]?((\d+(\.\d*)?)|\.\d+)([eE][\+-]?\d+)?\s*$"
# automata: http://images.cnitblog.com/i/627993/201405/012016243309923.png
class Solution:
# @param s, a string
# @return a boolean
def isNumber(self, s):
transition_table = [[-1, 0, 3, 1, 2, -1], # next states for state 0
[-1, 8, -1, 1, 4, 5], # next states for state 1
[-1, -1, -1, 4, -1, -1], # next states for state 2
[-1, -1, -1, 1, 2, -1], # next states for state 3
[-1, 8, -1, 4, -1, 5], # next states for state 4
[-1, -1, 6, 7, -1, -1], # next states for state 5
[-1, -1, -1, 7, -1, -1], # next states for state 6
[-1, 8, -1, 7, -1, -1], # next states for state 7
[-1, 8, -1, -1, -1, -1]] # next states for state 8
state = 0
for char in s:
inputType = InputType.INVALID
if char.isspace():
inputType = InputType.SPACE;
elif char == '+' or char == '-':
inputType = InputType.SIGN
elif char.isdigit():
inputType = InputType.DIGIT
elif char == '.':
inputType = InputType.DOT
elif char == 'e' or char == 'E':
inputType = InputType.EXPONENT;
state = transition_table[state][inputType];
if state == -1:
return False;
return state == 1 or state == 4 or state == 7 or state == 8
class Solution2:
# @param s, a string
# @return a boolean
def isNumber(self, s):
import re
return bool(re.match("^\s*[\+-]?((\d+(\.\d*)?)|\.\d+)([eE][\+-]?\d+)?\s*$", s))
if __name__ == "__main__":
print Solution().isNumber(" 0.1 ")
print Solution().isNumber("abc")
print Solution().isNumber("1 a")
print Solution().isNumber("2e10")