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.

class Solution {
public:
    int maximalRectangle(vector<vector<char> > &matrix) {
        int rows = matrix.size();
        if (rows == 0) return 0;
        int cols = matrix[0].size();
        if (cols == 0) return 0;
        cols++;
        
        int* memo = new int[cols * cols];
        for (int i=cols * cols - 1; i>=0; i--) memo[i] = 0;
        for (int i=1; i<cols; i++) memo[i * cols + i - 1] = 1;
        int max_rect = 0;
        for (int i=0; i<rows; i++) {
            vector<char>& row = matrix[i];
            for (int j = 1; j<cols; j++) {
                int base = j * cols;
                for (int k = j; k<cols; k++) {
                    if (row[k - 1] == '1' && memo[base + k - 1] > 0) {
                        int cur_rect = ++memo[base + k] * (k - j + 1);
                        if (cur_rect > max_rect) max_rect = cur_rect;
                    } else {
                        memo[base + k] = 0;
                    }
                }
            }
        }
        delete[] memo;
        return max_rect;
    }
};

O(n^3)暴力，用一个memo数组记录各个区段上的高度（memo[i+1][j+1]表示扫描到matrix中某一行时，该行[x_row][i...j]区间内的最低高度），用时250+ms，按照leetcode的一般时间范围，肯定有巧妙的解法，去zhuli哥看了一下，结合做max rectangular area in histogram那题可以把时间降到O（n^2），用时70+ms,缺乏灵活运用能力啊！下面给出代码

    int maximalRectangle(vector<vector<char> > &matrix) {
        int rows = matrix.size();
        if (rows == 0) return 0;
        int cols = matrix[0].size();
        if (cols == 0) return 0;

        vector<int> height(cols, 0);
        vector<int> L, R;
        L.resize(cols), R.resize(cols);
        
        int max_rect = 0;
        
        for (int i=0; i<rows; i++) {
            for (int j=0; j<cols; j++) {
                height[j] = matrix[i][j] == '1' ? height[j] + 1 : 0;
            }
            for (int j=0; j<cols; j++) {
                L[j] = j;
                while (L[j] - 1 >= 0 && height[L[j] - 1] >= height[j]) {
                    L[j] = L[L[j] - 1];
                }
            }
            for (int j=cols-1; j>=0; j--) {
                R[j] = j;
                while (R[j] + 1 < cols && height[R[j] + 1] >= height[j]) {
                    R[j] = R[R[j] + 1];
                }
            }
            for (int j=0; j<cols; j++) {
                int rect = (R[j] - L[j] + 1) * height[j];
                if (rect > max_rect) max_rect = rect;
            }
        }
        return max_rect;
    }

接着用那题maximum rectangle in histogram中stack版本来实现：

class Solution {
public:
    int maximalRectangle(vector<vector<char>>& matrix) {
        int area = 0;
        int rows = matrix.size();
        if (rows < 1) {
            return area;
        }
        int cols = matrix[0].size();
        if (cols < 1) {
            return cols;
        }
        vector<int> h(cols + 1, 0);
        
        for (int i=0; i<rows; i++) {
            for (int j=0; j<cols; j++) {
                h[j] = matrix[i][j] == '0' ? 0 : (h[j] + 1);
            }
            area = max(area, max_rect(h));
        }
        return area;
    }
    
    int max_rect(vector<int>& height) {
        int len = height.size();
        stack<int> pos;
        int maxarea = 0;
        for (int i=0; i<len; i++) {
            while (!pos.empty() && height[i] < height[pos.top()]) {
                int h = height[pos.top()];
                pos.pop();
                int w = i - (pos.empty() ? -1 : pos.top()) - 1;
                maxarea = max(maxarea, h * w);
            }
            pos.push(i);
        }
        return maxarea;
    }
};

参考：

　　http://www.cnblogs.com/zhuli19901106/p/3570175.html