題目描述:

給定一個只包括 ‘(‘，‘)‘，‘{‘，‘}‘，‘[‘，‘]‘ 的字符串，判斷字符串是否有效。

有效字符串需滿足：

1. 左括號必須用相同類型的右括號閉合。
2. 左括號必須以正確的順序閉合。

註意空字符串可被認為是有效字符串。

方法1：

　　遇到左括號壓入list中，遇到右括號時先判斷list是否為空，是則返回false，否則彈出一個字符與其進行比較，匹配則continue 否則返回false (32ms)

 1 class Solution(object):
 2     def isValid(self, s):
 3         """
 4         :type s: str
 
 5         :rtype: bool
 6         """
 7         lists = []
 8         i = 0
 9         if len(s) == 1:
10             return False
11         elif len(s) == 0:
12             return True
13         while i < len(s):
14             c = s[i]
15             if c ==‘(‘ or c == ‘[‘ or c == ‘{‘:
16                #
 
 if i + 1 != len(s):
17                     # if s[i+1] != ‘)‘ and s[i+1] != ‘]‘ and s[i+1] != ‘}‘:
18                     #     if c < s[i+1]:
19                     #         return False
20                     #([])這種竟然算對，好吧
21                 lists.append(c)
22                 
23             elif
 
 c == ‘)‘:
24 
25                 if len(lists) != 0:
26                     t = lists.pop()
27                     if t == ‘(‘:
28                         i+=1
29                         continue
30                     else:
31                         return False
32                 else:
33                     return False
34             elif c == ‘]‘:
35                
36                 if len(lists) != 0:
37                     t = lists.pop()
38                     if t == ‘[‘:
39                         i+=1
40                         continue
41                     else:
42                         return False
43                 else:
44                     return False
45             elif c == ‘}‘:
46                 
47                 if len(lists) != 0:
48                     t = lists.pop()
49                     if t == ‘{‘:
50                         i+=1
51                         continue
52                     else:
53                         return False
54                 else:
55                     return False
56             i += 1
57
58         if len(lists) != 0:
59             return False
60         else:
61             return True

簡潔版:(28ms)

 1 class Solution:
 2     def isValid(self, s):
 3         """
 4         :type s: str
 5         :rtype: bool
 6         """
 7         stack = []
 8         for c in s:
 9             if c == ‘(‘ or c == ‘{‘ or c == ‘[‘:
10                 stack.append(c)
11             elif not stack:
12                 return False
13             elif c == ‘)‘ and stack.pop() != ‘(‘:
14                 return False
15             elif c == ‘}‘ and stack.pop() != ‘{‘:
16                 return False
17             elif c == ‘]‘ and stack.pop() != ‘[‘:
18                 return False
19         return not stack

利用字典:(24ms)

 1 class Solution(object):
 2     def isValid(self, s):
 3         """
 4         :type s: str
 5         :rtype: bool
 6         """
 7         pars = [None]
 8         parmap = {‘)‘: ‘(‘, ‘}‘: ‘{‘, ‘]‘: ‘[‘}
 9         for c in s:
10             if c in parmap:
11                 if parmap[c] != pars.pop():
12                     return False
13             else:
14                 pars.append(c)
15         return len(pars) == 1

時間最短:(20ms)

　　用拋出異常的方式把類似)()剔除

 1 class Solution(object):
 2     def isValid(self, s):
 3         """
 4         :type s: str
 5         :rtype: bool
 6         """
 7         try:
 8             stack = []
 9             for key in s :
10                 if(key == ‘(‘ or key == ‘[‘ or key == ‘{‘):
11                     stack.append(key)
12                 else:
13                     if((key == ‘)‘ and stack.pop() == ‘(‘) or
14                             (key == ‘]‘ and stack.pop() == ‘[‘) or
15                             (key == ‘}‘ and stack.pop() == ‘{‘)):
16                         pass
17                     else:
18                         return False
19             if(len(stack) == 0):
20                 return True
21         except IndexError:
22             return False
23         return False

2018-07-22 18:48:45

LeetCode--020--括號匹配