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.

思路：

该题可以看作Largest Rectangle in Histogram的变形，对于每行，求出一个height数组，然后求出可以得到的最大值。

参考资料：http://www.cnblogs.com/lichen782/p/leetcode_maximal_rectangle.html

代码：

    int maxRectangleInHistogram(vector<int> &height){
        stack<int> Stack;
        int i = 0, num = height.size();
        int result = 0;
        while(i < num){
            if(Stack.empty() || height[Stack.top()] <= height[i]){
                Stack.push(i++);
            }
            else{
                int tmp = Stack.top();
                Stack.pop();
                int area = height[tmp]*(Stack.empty()?i:(i-Stack.top()-1));
                if(area > result)
                    result = area;
            }
        }
        while(!Stack.empty()){
            int tmp = Stack.top();
            Stack.pop();
            int area = height[tmp]*(Stack.empty()?i:(i-Stack.top()-1));
            if(area > result)
                result = area; 
        }
        return result;
    }
    int maximalRectangle(vector<vector<char> > &matrix) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int m = matrix.size();
        if(m == 0)
            return 0;
        int n = matrix[0].size();
        if(n == 0)
            return 0;
        vector<vector<int> > height(m, vector<int>(n, 0));
        int i,j;
        for(i = 0; i < m; i++){
            for(j = 0; j < n; j++){
                if(matrix[i][j] == '0')
                    height[i][j] = 0;
                else
                    height[i][j] = i==0 ? 1 : height[i-1][j]+1;
            }
        }
        int result = 0;
        for(i = 0; i < m; i++){
            int tmp = maxRectangleInHistogram(height[i]);
            if(tmp > result)
                result = tmp; 
        }
        return result;
    }