题目描述:Valid 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 "(]&qu…

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 "(]" and "([)]"…

题目在这里: https://leetcode.com/problems/valid-parentheses/ [标签]Stack; String [个人分析]这个题应该算是Stack的经典应用.先进后出 ( FILO) 的结构: 先来的左边括号跟后面的右边括号相匹配. [代码解释]创建一个栈,如果遇到的是 '{', '(', '[', 就压到栈里面去,留着跟后面遇到的后边括号匹配.如果遇到了'}', ']', ')',首先看看栈里面 是不是空,里面有没有匹配的部分. [一些心得]我在别人那儿看…

Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring. For "(()", the longest valid parentheses substring is "()", which has length = 2. Another example is &…

Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring. For "(()", the longest valid parentheses substring is "()", which has length = 2. Another example is &…

Given a string containing just the characters'('and')', find the length of the longest valid (well-formed) parentheses substring. For"(()", the longest valid parentheses substring is"()", which has length = 2. Another example is")…

(Version 1.3) 这题在LeetCode上的标签比较有欺骗性,虽然标签写着有DP,但是实际上根本不需要使用动态规划,相反的,使用动态规划反而会在LeetCode OJ上面超时.这题正确的做法应该和Largest Rectangle in Histogram那几个使用stack来记录并寻找左边界的题比较类似,因为在仔细分析问题并上手尝试解决时,会发现问题的关键在于怎么判定一个valid parentheses子串的起始位置,或者说当遇到一个')'时,怎么知道要加到哪里去. 第一次做的时候…

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. Open brackets must be closed in the correct…

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. Open brackets must be closed in the correct…

第一种方法,用栈实现,最容易想到,也比较容易实现,每次碰到‘)’时update max_len,由于要保存之前的‘(’的index,所以space complexity 是O(n) // 使用栈,时间复杂度 O(n),空间复杂度 O(n) class Solution { public: int longestValidParentheses(string s) { , last = -; stack<int> lefts; ; i < s.size(); ++i) { if (s[i]…