leetcode -- Valid Parentheses -- 简单重点

https://leetcode.com/problems/valid-parentheses/

要用到stack。 参考http://www.cnblogs.com/zuoyuan/p/3779772.html

正确code:

class Solution:
    # @return a boolean
    def isValid(self, s):
        stack = []
        for i in range(len(s)):
            if s[i] == '(' or s[i] == '[' or s[i] == '{':
                stack.append(s[i])
            if s[i] == ')':
                if stack == [] or stack.pop() != '(':
                    return False
            if s[i] == ']':
                if stack == [] or stack.pop() != '[':
                    return False
            if s[i] == '}':
                if stack == [] or stack.pop() != '{':
                    return False
        if stack:
            return False
        else:
            return True

my code:(不能过leetcode，供自己总结)
思路是用dict，再来判断是否match。但”([)]”这种case过不了，所以题目要求应该是括号严格match，左括号与右括号之间不能有其他括号。

class Solution(object):
    def isValid(self, s):
        """ :type s: str :rtype: bool """
        help = {'(':')','[':']','{':'}'}
        if not s: return True
        mydict = {}

        for i in xrange(len(s)):
            if s[i] in mydict:
                mydict[s[i]].append(i)
            else:
                mydict[s[i]] = [i]

        for k,v in help.items():

            if (k in mydict) and (v not in mydict):
                return False
            elif (v in mydict) and (k not in mydict):
                return False
            elif (v not in mydict) and (k not in mydict):
                continue
            else:
                list1 = mydict[k]
                list2 = mydict[v]
                if len(list1) != len(list2):
                    return False
                elif min(list2) <= min(list1) or max(list2) <= max(list1):
                    return False

        return True