Valid Parentheses

时间：2014-10-27 14:17:47 收藏：0 阅读：186
典型的用栈（stack）结构解决问题的例子
class Solution:
    def isPair(self,s1,s2):
        if (s1==‘(‘ and s2==‘)‘)or (s2==‘(‘ and s1==‘)‘):
            return True
        if (s1==‘[‘ and s2==‘]‘)or (s2==‘[‘ and s1==‘]‘):
            return True
        if (s1==‘{‘ and s2==‘}‘)or (s2==‘{‘ and s1==‘}‘):
            return True
        else :
            return False
            
    def isValid(self, s):

        if s=="":
            return True
        
        pair_right=set([‘)‘,‘]‘,‘}‘])
        if s[0] in pair_right :
            return False

        len_s=len(s)
        if len_s%2==1:
            return False
        
        i=1;
        stack=[s[0]]
        while(i<len_s):
            stack.append(s[i])
            len_stack=len(stack)
            if len_stack>1:
                if self.isPair(stack[len_stack-1],stack[len_stack-2]):
                    stack.pop()
                    stack.pop()
                else:
                    pass
            i+=1
        len_stack=len(stack)
        if len(stack)==0:
            return True
        else:
            return False