leetcode1106

Posted 2022-03-01 asenyang

做了一个解，但是WA，54/70，这个无法通过的测试用例有300个字符，太复杂了，我现在无法通过测试用例分析出哪里写的不对，感觉换一个思路更好一些：

 1 class Solution:
 2     def parseBoolExpr(self, expression: str) -> bool:
 3         n = len(expression)
 4         opstack = []
 5         boolstack = []
 6         savenum = []
 7         curnum = 0
 8         for i in range(n):
 9             cur = expression[i]
10             if cur == ‘(‘:
11                 if curnum > 0:
12                     savenum.append(curnum)
13                     curnum = 0
14             elif cur == ‘)‘:
15                 op = opstack.pop(-1)
16                 if op == ‘!‘:
17                     boolstack[-1] = not boolstack[-1]
18                 else:
19                     if op == ‘&‘:
20                         bsop = True
21                         bn = len(boolstack)
22                         blist = boolstack[bn-curnum:]
23                         while curnum > 0:
24                             boolstack.pop()
25                             curnum -= 1
26                         for b in blist:
27                             bsop = bsop and b
28                         boolstack.append(bsop)
29                     else:
30                         bsop = False
31                         bn = len(boolstack)
32                         blist = boolstack[bn-curnum:]
33                         while curnum > 0:
34                             boolstack.pop()
35                             curnum -= 1
36                         for b in blist:
37                             bsop = bsop or b
38                         boolstack.append(bsop)
39                     if len(savenum) > 0:
40                         pnum = savenum.pop(-1)
41                         curnum = pnum + 1
42                     else:
43                         curnum = 1
44             elif cur == ‘&‘ or cur == ‘|‘ or cur == ‘!‘:
45                 opstack.append(cur)
46             elif cur == ‘t‘ or cur == ‘f‘:
47                 curnum += 1
48                 bl = True if cur == ‘t‘ else False
49                 boolstack.append(bl)
50             elif cur == ‘,‘:
51                 continue
52             else:
53                 print(‘error‘)
54         return boolstack.pop(-1)

 

下面给以AC的解，这个和我的思路接近，使用栈来存储运算顺序，使用set()来进行括号内部的运算，可以减少手工运算的错误。

参考：https://leetcode.com/problems/parsing-a-boolean-expression/discuss/323390/Python-simple-solution-using-stack-easy-to-understand

 1 class Solution(object):
 2     def parseBoolExpr(self, expression):
 3         """
 4         :type expression: str
 5         :rtype: bool
 6         """
 7         stack = []
 8         for e in expression:
 9             if e in ‘!|&ft(‘:
10                 stack.append(e)
11             if e == ‘)‘:
12                 exp = set()
13                 while stack:
14                     b = stack.pop()
15                     if b != ‘(‘:
16                         exp.add(True if b == ‘t‘ else False)
17                     else:
18                         operator = stack.pop()
19                         if operator == ‘&‘:
20                             r = all(exp) 
21                         if operator == ‘|‘:
22                             r = any(exp)
23                         if operator == ‘!‘:
24                             r = not all(exp)
25                         stack.append(‘t‘ if r else ‘f‘)
26                         break
27         return True if stack[-1] == ‘t‘ else False   

 

至于那种转换成eval()的嘛……