题目大意:给出一个只包含字符'('和')'的字符串S,求最长有效括号序列的长度. 很有趣的题目,有助于我们对这种人类自身制定的规则的深入理解,可能我们大多数人都从没有真正理解过怎样一个括号序列是有效的,因此解题也无从说起.整道题目的难度在于我们对有效括号序列的理解和定义.下面给出我自己的定义:. 定义1:空括号序列是有效的. 定义2:对于一对左右括号,若左括号出现在右括号的左边,且左右括号之间(不包含两端)的括号序列是有效的,那么称该左括号到该右括号(包含)这一段序列是有效的.且称该左括号和右括…
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")…
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]…
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 &…
Longest Valid Parentheses 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…
(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…
class Solution(object): def longestValidParentheses(self, s): """ :type s: str :rtype: int """ maxlen=0 stack=[] last=-1 for i in range(len(s)): if s[i] == '(': stack.append(i) else: if stack == []: last=i else: stack.pop() i…
指数:[LeetCode] Leetcode 指标解释 (C++/Java/Python/Sql) Github: https://github.com/illuz/leetcode 032. Longest Valid Parentheses (Hard) 链接: 题目:https://oj.leetcode.com/problems/longest-valid-parentheses/ 代码(github):https://github.com/illuz/leetcode 题意: 问一个字…