Leetcode#85 Maximal Rectangle

Original title Address

Specifying any row as the lower boundary of the rectangle, the problem becomes the rectangle that asks for the maximum area of the histogram, see previous article.
So the idea is, from top to bottom, in order to find the histogram of the maximum area of the rectangle, feeling instantaneous no difficulty. In the following code, the first function is moved in intact.

Time complexity is O (n^2)

Code:

1     intLargestrectanglearea (vector<int> &height) {2         intMaxarea =0;3         intn =height.size ();4stack<int>St;5         inth, W;6 7          for(inti =0; I < n; i++) {8             if(St.empty () | | height[st.top ()] <Height[i])9 St.push (i);Ten             Else { One                  while(!st.empty () && height[st.top ()] >Height[i]) { Ah =height[st.top ()]; - St.pop (); -W = St.empty ()? I:i-(St.top () +1); theMaxarea = Max (Maxarea, H *W); -                 } - St.push (i); -             } +         } -  +          while(!St.empty ()) { Ah =height[st.top ()]; at St.pop (); -W = St.empty ()? -N-(st.top () +1); -Maxarea = Max (Maxarea, H *W); -         } -  -         returnMaxarea; in     } -  to     intMaximalrectangle (vector<vector<Char> > &matrix) { +         if(Matrix.empty () | | matrix[0].empty ()) -             return 0; the  *         intMaxarea =0; $         inth =matrix.size ();Panax Notoginseng         intW = matrix[0].size (); -vector<int> Histo (w,0); the  +          for(inti =0; I < H; i++) { A             //Calculate Histogram the              for(intj =0; J < W; J + +) +HISTO[J] = (Histo[j] +1) * (Matrix[i][j]-'0'); -  $Maxarea =Max (Maxarea, Largestrectanglearea (Histo)); $         } -  -         returnMaxarea; the}

Leetcode#85 Maximal Rectangle