LeetCode224:基本计算器(栈)

 

 解题思路:

1、双栈模拟,一个用来存数,一个用来存操作符。需要考虑 '('后面紧跟'+'、'-'这种情况

2、递归:遇到左括号开始递归,遇到右括号结束递归,返回值。

 

 1 class Solution:
 2     def calculate(self, s):
 3         s = '('+s+')'
 4         stack = []
 5         op = []
 6         val = 0
 7         flag = False
 8         for i in range(len(s)):
 9             if ord('0')<=ord(s[i])<=ord('9'):
10                 val=val*10+ord(s[i])-ord('0')
11                 flag= True
12             elif s[i]=='(':
13                 op.append('(')
14                 stack.append('(')
15             elif s[i]==')':
16                 if flag:
17                     stack.append(val)
18                     flag = False
19                     val = 0
20                 ans = 0
21                 while op[-1]!='(' and stack[-1]!='(':
22                     if op[-1]=='+':
23                         ans = ans + stack[-1]
24                     elif op[-1]=='-':
25                         ans = ans - stack[-1]
26                     _,_= stack.pop(),op.pop()
27 
28                 if op[-1]=='(' and stack[-1]=='(':
29                     _ = op.pop(),stack.pop()
30                 elif op[-1]=='(':
31                     ans+=stack[-1]
32                     _ = stack.pop()
33                     _ = op.pop(), stack.pop()
34                 else:
35                     print('error')
36                     exit()
37                 stack.append(ans)
38             elif s[i]=='+':
39                 if flag:
40                     stack.append(val)
41                     flag=False
42                     val = 0
43                 op.append('+')
44             elif s[i]=='-':
45                 if flag:
46                     stack.append(val)
47                     flag =False
48                     val=0
49                 op.append('-')
50             elif s[i]==' ':
51                 continue
52         return stack[-1]
53 s = Solution().calculate("0+0+(-0+0)")
54 print(s)

 

posted @ 2021-03-10 17:14  ISGuXing  阅读(70)  评论(0编辑  收藏  举报