Valid Number
Created: September 5, 2018 by [lek-tin]
Last updated: September 5, 2018
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.
Solution (naive)
using builtin function in python
class Solution:
def isNumber(self, s: str) -> bool:
try:
float(s)
return True
except ValueError:
return False
Solution (using flags: without natural number e)
class Solution:
def isNumberWithoutE(self, s: str) -> bool:
s = s.strip()
N = len(s)
if N == 0:
return False
hasSeenDot, hasSeenNum = False, False
for i in range(len(s)):
currChar = s[i]
# Number
if ord("0") <= ord(currChar) <= ord("9"):
hasSeenNum = True
# dot
elif currChar == ".":
if hasSeenDot:
print(currChar,": invalid '.'")
return False
hasSeenDot = True
# +/- sign
elif currChar in "+-":
# +/- at invalid position
if i != 0:
print(currChar, ": +/- at invalid position")
return False
# illegal characters
else:
print(currChar, ": illegal characters")
return False
return hasSeenNum
Solution (using flags: with natural number e)
class Solution:
def isNumber(self, s: str) -> bool:
s = s.strip()
N = len(s)
if N == 0:
return False
hasSeenE, hasSeenDot, hasSeenNum, numAfterE = False, False, False, False
for i in range(len(s)):
currChar = s[i]
# Number
if ord("0") <= ord(currChar) <= ord("9"):
hasSeenNum = True
numAfterE = True
# e
elif currChar == "e":
if hasSeenE or not hasSeenNum:
print(currChar,": invalid 'e'")
return False
hasSeenE = True
numAfterE = False
# dot
elif currChar == ".":
if hasSeenE or hasSeenDot:
print(currChar,": invalid '.'")
return False
hasSeenDot = True
# +/- sign
elif currChar in "+-":
# +/- at invalid position
if i != 0 and s[i-1] != "e":
print(currChar, ": +/- at invalid position")
return False
# illegal characters
else:
print(currChar, ": illegal characters")
return False
return hasSeenNum and numAfterE
Solution (Deterministic finite automaton)
class Solution:
def isNumber(self, s: str) -> bool:
INVALID = 0
SPACE = 1
SIGN = 2
DIGIT = 3
DOT = 4
NATURAL_NUM = 5
# row: state; col: token
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
]
# Starting state
state = 0
for char in s:
if state==-1:
return False
if char==" ":
token = SPACE
elif char in "-+":
token = SIGN
elif char in map(str, range(10)):
token = DIGIT
elif char==".":
token = DOT
elif char in "Ee":
token = NATURAL_NUM
else:
token = INVALID
state = transition_table[state][token]
if state in (1, 4, 7, 8): # accept state
return True
else:
return False
