【leetcode系列】Valid Parentheses】的更多相关文章

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]…
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 ']', 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…
20. Valid Parentheses 错误解法: "[])"就会报错,没考虑到出现')'.']'.'}'时,stack为空的情况,这种情况也无法匹配 class Solution { public: bool isValid(string s) { if(s.empty()) return false; stack<char> st; st.push(s[]); ;i < s.size();i++){ if(s[i] == '(' || s[i] == '['…