题目如下:
A valid parentheses string is either empty
(""),"(" + A + ")", orA + B, whereAandBare valid parentheses strings, and+represents string concatenation. For example,"","()","(())()", and"(()(()))"are all valid parentheses strings.A valid parentheses string
Sis primitive if it is nonempty, and there does not exist a way to split it intoS = A+B, withAandBnonempty valid parentheses strings.Given a valid parentheses string
S, consider its primitive decomposition:S = P_1 + P_2 + ... + P_k, whereP_iare primitive valid parentheses strings.Return
Safter removing the outermost parentheses of every primitive string in the primitive decomposition ofS.
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 <= 10000S[i]is"("or")"Sis a valid parentheses string
解题思路:括号配对的题目在leetcode出现了很多次了,从左往右遍历数组,分别记录左括号和右括号出现的次数,当两者相等的时候,即为一组括号。
代码如下:
class Solution(object): def removeOuterParentheses(self, S): """ :type S: str :rtype: str """ left = 0 right = 0 res = ‘‘ tmp = ‘‘ for i in S: tmp += i if i == ‘(‘: left += 1 else: right += 1 if left == right: res += tmp[1:-1] tmp = ‘‘ left = 0 right = 0 return res
【leetcode】1021. Remove Outermost Parentheses
原文:https://www.cnblogs.com/seyjs/p/10673708.html