思路主要是:类似Valid Parentheses,用一个stack按顺序记录'('的index,再用一个变量记录当前可能的substring的开头。每当遇到一个')'时,若stack非空,则pop出一个元素,若pop之后非空,说明当前只能匹配到上一个尚未被匹配的'(',即stack.peek();若stack为空,说明可以一直匹配到start。当发现')'多于'('时,即...
[Leetcode] 20. Valid Parentheses(Stack) 括号匹配问题,使用栈的特点,匹配则出栈,否则入栈,最后栈为空则全部匹配。代码如下: 1classSolution {2public:3boolisValid(strings) {4stack<char>T;5for(inti =0;i < s.length();i ++)6{7if((T.empty()) || (s[i] == T.top()) || abs(s[i] ...
必须和新进来的右括号进行匹配,负责就是非法字符串## 这是一个比较巧妙的方法,先在栈底部加入一个元素,以解决空字符串的问题classSolution(object):defisValid(self,s):""":type s: str 字符串类型:rtype: bool 返回布尔型"""stack=['A']## 栈底加入了 A 字符m={')':'(',']':'[','...
if(stack.peek()!='{') returnfalse; stack.pop(); } } returnstack.size()==0; }
32. 最长有效括号 Longest Valid Parentheses难度:Hard| 困难 相关知识点:字符串 动态规划题目链接:https://leetcode-cn.com/problems/longest-valid-parentheses/官方题解:https://leetcode-cn.com/problems/longest-valid-parentheses/solution/, 视频播放量 3908、
用数组模拟stack, 用一个head指示栈头的下一个: 1publicclassSolution {2publicbooleanisValid(String s) {3char[] stack =newchar[s.length()];4inthead = 0;5for(charc : s.toCharArray()) {6switch(c) {7case'{':8case'[':9case'(':10stack[head++] =c;11break;12case'}':13if(head =...
【LeetCode 20. Valid Parentheses】(合法括号匹配判断,栈的应用),题目链接Givenastringcontainingjustthecharacters‘(‘,‘)’,‘{‘,‘}’,‘[’and‘]’,determineiftheinputstringisvalid.Aninputstringisvalidif:Openbracketsmustbeclosedbythesametyp...
Can you solve this real interview question? Valid Parentheses - Given a string s containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. An input string is valid if: 1. Open brackets must be closed by th
41. Longest Valid Parentheses 最长有效括号算法:本题用两种方法解,一是stack,二是双向遍历面试准备系列,注重培养互联网大厂的面试能力,注重编程思路,coding限时,代码规范,想要求职,转行或者跳槽的朋友们别忘了一键三连一下,新人up主需要你宝贵的支持哟~
20 Valid Parentheses 有效的括号 Description: Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. An input string is valid if: Open brackets must be closed by the same type of brackets. ...