82 lines
2.2 KiB
Python
82 lines
2.2 KiB
Python
import re
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from collections import namedtuple
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left = r'(?P<LEFT>\()'
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right = r'(?P<RIGHT>\))'
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word = r'(?P<WORD>[A-Z][a-z]*)'
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num = r'(?P<NUM>\d+)'
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pt = re.compile('|'.join([left, right, word, num]))
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token = namedtuple('token', ['type', 'value'])
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def genToken(s):
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scanner = pt.scanner(s)
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for i in iter(scanner.match, None):
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yield token(i.lastgroup, i.group(0))
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class parser:
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'''grammar:
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S-> item | S item
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item -> word | word num | '(' S ')' num
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'''
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def match(self, curType):
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sym = self.token[self.lookahead]
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if sym.type == curType:
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self.lookahead += 1
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return sym.value
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raise Exception('Invalid input string')
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def parse(self, s):
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self.token = [i for i in genToken(s)]
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self.lookahead = 0
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return self.S()
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def S(self):
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dic = {}
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while self.lookahead < len(self.token) and self.token[self.lookahead].type != 'RIGHT':
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cur = self.item()
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for i in cur:
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if i in dic:
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dic[i] += cur[i]
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else:
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dic[i] = cur[i]
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return dic
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def item(self):
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if self.token[self.lookahead].type == 'WORD':
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ele = self.match('WORD')
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n = 1
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if self.lookahead < len(self.token) and self.token[self.lookahead].type == 'NUM':
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n = int(self.match('NUM'))
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return {ele: n}
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elif self.token[self.lookahead].type == 'LEFT':
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self.match('LEFT')
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dic = self.S()
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self.match('RIGHT')
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n = int(self.match("NUM"))
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for i in dic:
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dic[i] *= n
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return dic
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else:
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print(self.token[self.lookahead])
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raise Exception('invalid string')
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class Solution(object):
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def countOfAtoms(self, formula):
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"""
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:type formula: str
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:rtype: str
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"""
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dic = parser().parse(formula)
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return ''.join(c+str(dic[c]) if dic[c] != 1 else c for c in sorted(dic.keys()))
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if __name__ == "__main__":
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li = ["K4(ON(SO3)2)2","Mg(OH)2"]
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sol = Solution()
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for s in li:
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print('>>>',s)
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print(sol.countOfAtoms(s))
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