108-Longest Valid Parentheses

-32. Longest Valid Parentheses My Submissions QuestionEditorial Solution
Total Accepted: 64569 Total Submissions: 285723 Difficulty: Hard
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.

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class Solution {public:    int longestValidParentheses(string s) {        int max_len=0,last = -1;        int n= s.size();        stack<int> lefts;        for(int i=0;i<n;++i){            if(s[i]=='(')lefts.push(i);//如果是左括号，一直入栈            else{                  if(lefts.empty()){//一直遇到右括号“)))))”,如果栈空，记录last的位置，                    //总是在第一个合法位置的前一个位置                    last = i;                   }                else{//遇到右括号且栈不为空，那么出栈且计算当前最大值                    lefts.pop();                    if(lefts.empty())//空了，记录上一次合法的第一个位置,序列")))))()"，出栈之前“(”                        max_len = max(max_len,i-last);                    else max_len = max(max_len,i-lefts.top());//这种情况，序列“((()”,出栈前“(((”                }            }        }        return max_len;    }    /**********************动态规划*******************************     *记f[i]为以i开头的合法括号长度最大值     *1.这种情况“（（）（）（））”f[i]=f[i+1]+2,     *如果s[i]与s[i+f[i+1]+1]匹配,match=i+f[i+1]+1     *即一个是'(',一个是')'     *2.如果是这样“（（）（）（））（）”,f[i]=f[i]+f[match+1]    ***********************************************************/    int longestValidParentheses_2(string s) {        int n = s.size();        vector<int> f(n,0);        int res=0;        for(int i=n-2;i>=0;--i){            int match = i+f[i+1]+1;            if(s[i]=='('&& match<n &&s[match]==')'){                f[i]=f[i+1]+2;                if(match+1<n)f[i]+=f[match+1];            }            res = max(res,f[i]);         }        return res;    }};