Ai
1 Star 2 Fork 5

LilithSangreal/LeetCode-Solutions

加入 Gitee
与超过 1200万 开发者一起发现、参与优秀开源项目,私有仓库也完全免费 :)
免费加入
文件
克隆/下载
valid-number.py 2.04 KB
一键复制 编辑 原始数据 按行查看 历史
Allen Liu 提交于 2018-12-30 21:04 +08:00 . remove semicolons
# Time: O(n)
# Space: O(1)
class InputType(object):
INVALID = 0
SPACE = 1
SIGN = 2
DIGIT = 3
DOT = 4
EXPONENT = 5
# regular expression: "^\s*[\+-]?((\d+(\.\d*)?)|\.\d+)([eE][\+-]?\d+)?\s*$"
# automata: http://images.cnitblog.com/i/627993/201405/012016243309923.png
class Solution(object):
def isNumber(self, s):
"""
:type s: str
:rtype: bool
"""
transition_table = [[-1, 0, 3, 1, 2, -1], # next states for state 0
[-1, 8, -1, 1, 4, 5], # next states for state 1
[-1, -1, -1, 4, -1, -1], # next states for state 2
[-1, -1, -1, 1, 2, -1], # next states for state 3
[-1, 8, -1, 4, -1, 5], # next states for state 4
[-1, -1, 6, 7, -1, -1], # next states for state 5
[-1, -1, -1, 7, -1, -1], # next states for state 6
[-1, 8, -1, 7, -1, -1], # next states for state 7
[-1, 8, -1, -1, -1, -1]] # next states for state 8
state = 0
for char in s:
inputType = InputType.INVALID
if char.isspace():
inputType = InputType.SPACE
elif char == '+' or char == '-':
inputType = InputType.SIGN
elif char.isdigit():
inputType = InputType.DIGIT
elif char == '.':
inputType = InputType.DOT
elif char == 'e' or char == 'E':
inputType = InputType.EXPONENT
state = transition_table[state][inputType]
if state == -1:
return False
return state == 1 or state == 4 or state == 7 or state == 8
class Solution2(object):
def isNumber(self, s):
"""
:type s: str
:rtype: bool
"""
import re
return bool(re.match("^\s*[\+-]?((\d+(\.\d*)?)|\.\d+)([eE][\+-]?\d+)?\s*$", s))
Loading...
马建仓 AI 助手
尝试更多
代码解读
代码找茬
代码优化
1
https://gitee.com/LilithSangreal/LeetCode-Solutions.git
git@gitee.com:LilithSangreal/LeetCode-Solutions.git
LilithSangreal
LeetCode-Solutions
LeetCode-Solutions
master

搜索帮助