LeetCode--020--括号匹配

题目描述:

给定一个只包括 '('，')'，'{'，'}'，'['，']' 的字符串，判断字符串是否有效。

有效字符串需满足：

1. 左括号必须用相同类型的右括号闭合。
2. 左括号必须以正确的顺序闭合。

注意空字符串可被认为是有效字符串。

方法1：

　　遇到左括号压入list中，遇到右括号时先判断list是否为空，是则返回false，否则弹出一个字符与其进行比较，匹配则continue 否则返回false   (32ms)

 class Solution(object):
     def isValid(self, s):
         """
         :type s: str
         :rtype: bool
         """
         lists = []
         i = 0
         if len(s) == 1:
             return False
         elif len(s) == 0:
             return True
         while i < len(s):
             c = s[i]
             if c =='(' or c == '[' or c == '{':
                # if i + 1 != len(s):
                     # if s[i+1] != ')' and s[i+1] != ']' and s[i+1] != '}':
                     #     if c < s[i+1]:
                     #         return False
                     #([])这种竟然算对，好吧
                 lists.append(c)

             elif c == ')':

                 if len(lists) != 0:
                     t = lists.pop()
                     if t == '(':
                         i+=1
                         continue
                     else:
                         return False
                 else:
                     return False
             elif c == ']':

                 if len(lists) != 0:
                     t = lists.pop()
                     if t == '[':
                         i+=1
                         continue
                     else:
                         return False
                 else:
                     return False
             elif c == '}':

                 if len(lists) != 0:
                     t = lists.pop()
                     if t == '{':
                         i+=1
                         continue
                     else:
                         return False
                 else:
                     return False
             i += 1

         if len(lists) != 0:
             return False
         else:
             return True

简洁版:(28ms)

 class Solution:
     def isValid(self, s):
         """
         :type s: str
         :rtype: bool
         """
         stack = []
         for c in s:
             if c == '(' or c == '{' or c == '[':
                 stack.append(c)
             elif not stack:
                 return False
             elif c == ')' and stack.pop() != '(':
                 return False
             elif c == '}' and stack.pop() != '{':
                 return False
             elif c == ']' and stack.pop() != '[':
                 return False
         return not stack

利用字典:(24ms)

 class Solution(object):
     def isValid(self, s):
         """
         :type s: str
         :rtype: bool
         """
         pars = [None]
         parmap = {')': '(', '}': '{', ']': '['}
         for c in s:
             if c in parmap:
                 if parmap[c] != pars.pop():
                     return False
             else:
                 pars.append(c)
         return len(pars) == 1

时间最短:(20ms)

　　用抛出异常的方式把类似)()剔除

 class Solution(object):
     def isValid(self, s):
         """
         :type s: str
         :rtype: bool
         """
         try:
             stack = []
             for key in s :
                 if(key == '(' or key == '[' or key == '{'):
                     stack.append(key)
                 else:
                     if((key == ')' and stack.pop() == '(') or
                             (key == ']' and stack.pop() == '[') or
                             (key == '}' and stack.pop() == '{')):
                         pass
                     else:
                         return False
             if(len(stack) == 0):
                 return True
         except IndexError:
             return False
         return False

2018-07-22 18:48:45