Feb_0219_Leetcode_32_Longest Valid Parentheses

Description

Given a string containing just the characters

'('

and

')'

, find the length of the longest valid (well-formed) parentheses substring.

For

"(()"

, the longest valid parentheses substring is

"()"

, which has length = 2.

Another example is

")()())"

, where the longest valid parentheses substring is

"()()"

, which has length = 4.

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Note:

The longest vaild parentheses is a substring of s

My Solution:

class Solution(object):
def longestValidParentheses(self, s):
if len(s) == 0:
return 0
s,stack,stack_num=')'+s,[')'],[0] #加入括号的原因防止栈为空
for i in range(1, len(s)):
if stack[-1]=='(' and s[i]==')':
stack.pop()
stack_num.pop()
else:
stack.append(s[i])
stack_num.append(i)
stack_num.append(len(s))
res=0
for j in range(len(stack_num)-1):
res=max(res,stack_num[j+1]-stack_num[j]-1)
return res

思路：

1. 创建两个栈，一个用来放字符，一个用来放字符对应的索引，为了防止栈为空分别加入‘)’，0,

2.遍历字符串，如果当前的字符串的字符为‘)’并且字符栈中的最上面为的字符为‘（’，这样意味着即将匹配成功，就把栈顶的字符pop出去（可以看成是先把当前的字符压栈，再pop出去一对），否则压栈，并把对应字符的索引压到另外一个数据栈。

3.生成一个索引的list，可以发现，留下来的数据之间和匹配成功的对数有关系，即两两之间的差减一，然后就可以找到最大值