Description
A valid parentheses string is either empty (""), "(" + A + ")", or A + B, where A and B are valid parentheses strings, and + represents string concatenation. For example, "", "()", "(())()", and "(()(()))" are all valid parentheses strings.
A valid parentheses string S is primitive if it is nonempty, and there does not exist a way to split it into S = A+B, with A and B nonempty valid parentheses strings.
Given a valid parentheses string S, consider its primitive decomposition: S = P_1 + P_2 + ... + P_k, where P_i are primitive valid parentheses strings.
Return S after removing the outermost parentheses of every primitive string in the primitive decomposition of S.
有效括号字符串为空 ("")、"(" + A + ")" 或 A + B，其中 A 和 B 都是有效的括号字符串，+ 代表字符串的连接。例如，""，"()"，"(())()" 和 "(()(()))" 都是有效的括号字符串。
如果有效字符串 S 非空，且不存在将其拆分为 S = A+B 的方法，我们称其为原语（primitive），其中 A 和 B 都是非空有效括号字符串。
给出一个非空有效字符串 S，考虑将其进行原语化分解，使得：S = P_1 + P_2 + ... + P_k，其中 P_i 是有效括号字符串原语。
对 S 进行原语化分解，删除分解中每个原语字符串的最外层括号，返回 S 。
题目链接：https://leetcode.com/problems/remove-outermost-parentheses/
Difficulty: easy
Example 1:
Input: "(()())(())"
Output: "()()()"
Explanation:
The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
After removing outer parentheses of each part, this is "()()" + "()" = "()()()".
Example 2:
Input: "(()())(())(()(()))"
Output: "()()()()(())"
Explanation:
The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".
After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".
Example 3:
Input: "()()"
Output: ""
Explanation:
The input string is "()()", with primitive decomposition "()" + "()".
After removing outer parentheses of each part, this is "" + "" = "".
Note:
- S.length <= 10000
- S[i] is “(“ or “)”
- S is a valid parentheses string
分析
- 用index记录匹配的左右括号的数目，负数代表左括号“（”比右括号“）”多，绝对值代表多的数目；
- 遍历S，res存储删除最外层括号的结果，若index=0且当前字符为“（”，index-=1，继续；
- 若index=-1，且当前字符为“）”，index+=1，继续，（这是匹配到最外层括号的情况）；
- 反之当前字符加到res中，若当前字符为“（”，index-=1，反之，index+=1；
- 返回res。
参考代码
class Solution(object):
def removeOuterParentheses(self, S):
res=''
index=0
for s in S:
if(index==0 and s=='('):
index-=1
continue
elif(index==-1 and s==')'):
index+=1
continue
else:
res+=s
if(s=='('):
index-=1
else:
index+=1
return res