921. Minimum Add to Make Parentheses Valid】的更多相关文章

Given a string S of '(' and ')' parentheses, we add the minimum number of parentheses ( '(' or ')', and in any positions ) so that the resulting parentheses string is valid. Formally, a parentheses string is valid if and only if: It is the empty stri…
Given a string S of '(' and ')' parentheses, we add the minimum number of parentheses ( '(' or ')', and in any positions ) so that the resulting parentheses string is valid. Formally, a parentheses string is valid if and only if: It is the empty stri…
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 日期 题目地址: https://leetcode.com/problems/minimum-add-to-make-parentheses-valid/ 题目描述 Given a string S of '(' and ')' parentheses, we add the minimum number of parentheses ( '(' o…
题目如下: 解题思路:上周都在忙着参加CTF,没时间做题,今天来更新一下博客吧.括号问题在leetcode中出现了很多,本题的解题思路和以前的括号问题一样,使用栈.遍历Input,如果是'('直接入栈:如果是')'则判断栈顶是否为'(',如果栈顶是'(',栈顶元素出栈,否则要加括号的count加1.遍历完成后,count + 栈里剩余的 '(' 的数量就是结果. 代码如下: class Solution(object): def minAddToMakeValid(self, S): "&quo…
921. 使括号有效的最少添加 921. Minimum Add to Make Parentheses Valid 题目描述 给定一个由 '(' 和 ')' 括号组成的字符串 S,我们需要添加最少的括号( '(' 或是 ')',可以在任何位置),以使得到的括号字符串有效. 从形式上讲,只有满足下面几点之一,括号字符串才是有效的: 它是一个空字符串,或者 它可以被写成 AB (A 与 B 连接),其中 A 和 B 都是有效字符串,或者 它可以被写作 (A),其中 A 是有效字符串. 给定一个括号…
Given a string S of '(' and ')' parentheses, we add the minimum number of parentheses ( '(' or ')', and in any positions ) so that the resulting parentheses string is valid. Formally, a parentheses string is valid if and only if: It is the empty stri…
Given a string S of '(' and ')' parentheses, we add the minimum number of parentheses ( '(' or ')', and in any positions ) so that the resulting parentheses string is valid. Formally, a parentheses string is valid if and only if: It is the empty stri…
最近遇到一个奇葩的问题,好郁闷 之前也没有仔细看.问题偶尔出现一次.再去查看日志时,出现 view.WindowManager$BadTokenException: Unable to add window…is not valid; is your activity running? 什么情况,activity is runing? 对呀,activity没有destory呀,在跑呢,怎么会出现 对话框 附加到activity上加不了呢,还说无效的,似是activity被destory了呢.…
Generate Parentheses Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]"…
给一组括号,remove最少的括号使得它valid 从左从右各scan一次 package fb; public class removeParen { public static String fix(String str) { StringBuffer res = new StringBuffer(str); int l = 0, r = 0; int i = 0; while (i < res.length()) { if (res.charAt(i) == '(') l++; else…