[LeetCode] Maximal Rectangle

Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing all ones and return its area.

在做Largest Rectangle in Histogram的时候有人说可以用在这题，看了一下还真是，以每行为x轴，每列往上累计的连续的1当成高度，就可以完全使用一样的方法了。

 int largestArea(vector<int>height){
     stack<int> s;
     int maxArea = ;
     int i = ;
     height.push_back();
     int len = height.size();

     while (i < len){
         if (s.empty() || height[s.top()] < height[i]){
             s.push(i++);
         }else{
             int t = s.top();
             s.pop();
             maxArea = max(maxArea, height[t] * (s.empty()? i : (i-s.top()-)));
         }
     }
     return maxArea;
 }

 int maximalRectangle(vector<vector<char>> &matrix){
     vector<int> height;
     int maxRect=;
     for (int row=; row<matrix.size(); row++){
         for (int col=; col<matrix[row].size(); col++){
             if (matrix[row][col] == ''){
                 height.push_back();
             }
             else{
                 int c=;
                 for (int i=row; i>-; i--){
                     if (matrix[i][col] != ''){
                         c++;
                     }else {
                         break;
                     }
                 }
                 height.push_back(c);
             }
         }

         for (int i=;i<height.size(); i++){
             cout << height[i] << " ";
         }
         cout << endl;

         maxRect = max(maxRect, largestArea(height));
         height.clear();
     }
     return maxRect;
 }

 int maximalRectangle2(vector<vector<char>> &matrix){
     int maxRect = ;
     if (matrix.size() < ) return ;
     vector<int>height(matrix[].size(), );
     for (int i=; i<matrix.size(); i++){
         for (int j=; j<matrix[i].size(); j++){
             if (matrix[i][j] == ''){
                 height[j] += ;
             }else{
                 height[j] = ;
             }
         }
         maxRect = max(maxRect, largestArea(height));
     }
     return maxRect;
 }

第一个maximalRectangle函数我用了很蠢的遍历方法来获得每行的height，这样活生生的把原本O(n^2)搞成了O(n^3)。。。 最后看了别人博客，说height是可以通过前一行的值算出了的(有点类似DP的思想...如果这也算的话)，豁然开朗，才写出了

maximalRectangle2这个真正的O(n^2)方法。