题目如下:
A valid parentheses string is either empty
("")
,"(" + A + ")"
, orA + B
, whereA
andB
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 intoS = A+B
, withA
andB
nonempty valid parentheses strings.Given a valid parentheses string
S
, consider its primitive decomposition:S = P_1 + P_2 + ... + P_k
, whereP_i
are primitive valid parentheses strings.Return
S
after 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 <= 10000
S[i]
is"("
or")"
S
is 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