词法分析、语法分析专题
前言:这是字符串解析的大杀器,基本上能推出「上下文无关文法」就能做出这道题
0X00 模板
import re
from collections import namedtuple
token = namedtuple('token', ['type', 'value'])
left = r'(?P<LEFT>\{)'
right = r'(?P<RIGHT>\})'
word = r'(?P<WORD>[a-z]+)'
comma = r'(?P<COMMA>\,)'
blank = r'(?P<BLANK>\s)'
pt = re.compile('|'.join([left, right, word, comma, blank]))
def genToken(s):
scanner = pt.scanner(s)
for i in iter(scanner.match, None):
if i.lastgroup != 'BLANK':
yield token(i.lastgroup, i.group(0))
class parser:
'''gramar
expr -> item | item ',' expr
item -> factor | factor item
factor -> WORD | '{' expr '}'
'''
def match(self, tp):
# print(self.p.value)
if tp == self.p.type:
val = self.p.value
try:
self.p = next(self.gen)
except StopIteration:
self.p = None
except Exception as e:
print(e)
return val
else:
raise Exception(f"[Error]: {tp} expected, got {self.p.type}")
def parse(self, s):
self.gen = genToken(s)
self.p = next(self.gen)
st = self.expr()
return sorted(list(st))
def expr(self):
ret = self.item()
while self.p and self.p.type == 'COMMA':
self.match('COMMA')
ret = ret.union(self.item())
return ret
def item(self):
ret = self.factor()
while self.p and self.p.type in ['WORD', 'LEFT']:
sufs = self.factor()
new = set()
for pre in ret:
for suf in sufs:
new.add(pre+suf)
ret = new
return ret
def factor(self):
if self.p.type == 'LEFT':
self.match('LEFT')
ret = self.expr()
self.match('RIGHT')
return ret
return {self.match('WORD')}
class Solution:
def braceExpansionII(self, expression):
return parser().parse(expression)
if __name__ == '__main__':
sol = Solution()
li = ["{a,b}{c{d,e}}", "{{a,z}, a{b,c}, {ab,z}}", "{a,b}c{d,e}f"]
for i in li:
print('>>>', i)
print(sol.braceExpansionII(i))
我基本上写完这一套要 20min 哭了。。。
对于「上下文无关文法」的推导的关键是凑反正从 expr 开始,一般两三层可以搞定。
0X01 相关题目
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Lisp 语法解析(这个太难)
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