085 Maximal Rectangle [Leetcode]

题目内容：

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

DP的过程还没有想清楚，先利用上一题的结果可以解决该问题。统计每一行为止向上有多少个连续的1做为bar的高度，对每一行调用Largest Rectangle in Histogram即可。

代码如下，运行时间32ms：

class Solution {public:    int maximalRectangle(vector<vector<char>>& matrix) {        if(matrix.size() == 0 || matrix[0].size() == 0)            return 0;        int row(matrix.size()), col(matrix[0].size()), result(0);        vector<vector<int>> heights(row, vector<int>(col));        for(int i = 0; i < col; ++i)            heights[0][i] = matrix[0][i] - '0';        result = largestRectangleArea(heights[0]);        for(int i = 1; i < row; ++i) {            for(int j = 0; j < col; ++j) {                heights[i][j] = (matrix[i][j] - '0') ? heights[i-1][j] + 1 : 0;            }            int temp(largestRectangleArea(heights[i]));            result = result > temp ? result : temp;        }        return result;    }    int largestRectangleArea(vector<int>& height) {        if(height.size() == 0)             return 0;        //put a minimum number to calculate the area of last rectangle        height.push_back(0);        stack<int> h;        int result(0), index(0), top(0);        while(index < height.size()) {            if(h.empty() || height[h.top()] < height[index]) {                h.push(index++);            }            else {                top = h.top();                h.pop();                //if stack had only 1 element, use index as width, else use current top index to calculate width                top = h.empty() ? height[top] * index : height[top] * (index - h.top() - 1);                result = result > top ? result : top;            }        }        height.pop_back();        return result;    }};