problem 1021. Remove Outermost Parentheses 参考 1. Leetcode_easy_1021. Remove Outermost Parentheses; 完…

题目如下: 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, "", "()", "(())()"…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 遍历 日期 题目地址:https://leetcode.com/problems/remove-outermost-parentheses/ 题目描述 A valid parentheses string is either empty (""), "(" + A + ")", or A + B, w…

problem 1047. Remove All Adjacent Duplicates In String 参考 1. Leetcode_easy_1047. Remove All Adjacent Duplicates In String; 完…

网址:https://leetcode.com/problems/remove-outermost-parentheses/ 使用栈的思想,选择合适的判断时机 class Solution { public: string removeOuterParentheses(string S) { stack<char> s_ch; string ans; stack<char> temp; , r = ; ; i<S.size(); i++) { if(S[i] == '(')…

括号匹配想到用栈来做: class Solution: def removeOuterParentheses(self, S: str) -> str: size=len(S) if size==0: return "" i=0 strings=[] while i<size: stack=[S[i]] string=S[i] i+=1 while i<size: if S[i]=='(': stack.append(S[i]) else: stack.pop() s…

题目如下: 解题思路:还是这点经验,对于需要输出整个结果集的题目,对性能要求都不会太高.括号问题的解法也很简单,从头开始遍历输入字符串并对左右括号进行计数,其中出现右括号数量大于左括号数量的情况,表示这个区间是不合法的,需要删掉一个右括号:遍历完成后,如果左括号数量大于右括号的数量,那么需要删除左括号,直至两者相等. 代码如下: class Solution(object): def removeInvalidParentheses(self, s): """ :type s…

https://leetcode-cn.com/problems/remove-outermost-parentheses/ Java Solution class Solution { public String removeOuterParentheses(String S) { char[] chars = S.toCharArray(); int flag = 0; StringBuilder result = new StringBuilder(); for (char c : cha…

1021. 删除最外层的括号 1021. Remove Outermost Parentheses 题目描述 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…

709. To Lower Case(Easy)# Implement function ToLowerCase() that has a string parameter str, and returns the same string in lowercase. Example 1: Input: "Hello" Output: "hello" Example 2: Input: "here" Output: "here"…