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.

Largest Rectangle in Histogram的进阶篇，计算柱状图最大矩形的解法可以进一步优化 - 不用两元组堆栈，用一个索引堆栈即可。关键是找到左右两侧比当前处理元素小的索引，面积为当前元素高度和左右索引差值的乘积。其他的就水到渠成了，把矩阵的每一行之上的所有元素看做一个柱状图。

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