Longest Valid parentheses

Eliminate all valid parentheses, that is, a pair (), and use buyers to save the invalid parentheses in a single (or) subscript in a string. Then the difference between each of the two illegal subscript minus 1 is the length of the legal string in the middle, followed by a backward comparison, to select the largest string.

1 classSolution {2  Public:3     intLongestvalidparentheses (strings) {4stack<int>St;5      intN=s.length (), a,b,longest=0;6       for(intI=0; i<n;i++)7      {8          if(s[i]=='(') St.push (i);9          Else if(!st.empty ())//When the buyers is empty, its top is 0, so determine if it is emptyTen          { One           if(S[st.top ()]=='(') St.pop (); A              ElseSt.push (i); -          } -          ElseSt.push (i); the           -      } -A=N; -b=0; +      if(St.empty ()) longest=N; -      Else  +      { A           while(!st.empty ()) at          { -b=st.top (); St.pop (); -Longest=max (longest,a-b-1); -A=b; -          } -longest=Max (longest,a); in      } -      returnlongest; to     } +};

The following is a dynamic programming approach:

Each longest represents the most effective length of this segment, which is divided by ') '

1 classSolution {2  Public:3     intLongestvalidparentheses (strings) {4         intn=s.length ();5         intMaxvalid=0;6vector<int> Longest (N,0);7        for(intI=1; i<n;i++)8       {9           if(s[i]==')')Ten           { One               if(s[i-1]=='(') A               { -longest[i]= (I-2) >=0? (longest[i-2]+2):2; -Maxvalid=Max (maxvalid,longest[i]); the               } -               Else if((i-longest[i-1]-1) >=0&&s[i-longest[i-1]-1]=='('//i-longest[i-1]-1 is to seek this ' (' corresponding ' (' position -               { -longest[i]=longest[i-1]+2+ ((i-longest[i-1]-2>=0)? longest[i-longest[i-1]-2]:0); Note that the last curly brace cannot be lost!!  +Maxvalid=Max (maxvalid,longest[i]); -               } +           } A       } at       returnMaxvalid; -     } -};

Longest Valid parentheses