首页 > 代码库 > [leetcode-32-Longest Valid Parentheses]
[leetcode-32-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 = 2.
Another example is ")()())"
, where the longest valid parentheses substring is "()()"
, which has length = 4.
思路
The workflow of the solution is as below.
- Scan the string from beginning to end.
- If current character is ‘(‘,
push its index to the stack. If current character is ‘)‘ and the
character at the index of the top of stack is ‘(‘, we just find a
matching pair so pop from the stack. Otherwise, we push the index of
‘)‘ to the stack. - After the scan is done, the stack will only
contain the indices of characters which cannot be matched. Then
let‘s use the opposite side - substring between adjacent indices
should be valid parentheses. - If the stack is empty, the whole input
string is valid. Otherwise, we can scan the stack to get longest
valid substring as described in step 3.
int longestValidParentheses(string s) { stack<int>st; if(s.length()<=1)return 0; for(int i=0;i<s.length();i++) { if(st.empty())st.push(i); else { if(s[i] ==‘(‘)st.push(i); else if(s[i]==‘)‘&&s[st.top()]==‘(‘)st.pop(); else st.push(i); } } if(st.empty()) return s.length(); int a =s.length(),b =0; int longest = 0; while(!st.empty()) { b=st.top(); st.pop(); longest = max(longest,a-b-1); a = b; } longest = max(longest,a); return longest; }
参考:
https://discuss.leetcode.com/topic/2289/my-o-n-solution-using-a-stack
[leetcode-32-Longest Valid Parentheses]
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。