* leetcode #85 in cpp

Solution: 

This question is hard. I did not know how to solve this until I looked up the solution online. Basically to solve this question we employ techniques in #84. That is, we treat this problem as looking for largest rectangle in histogram. 

Example:

given 

011111101We treat each row as the x-axis, and the 1s as the height of the histogram.  

For row 1:  it is [0, 1,1], max rectangle in this histogram is 2.

For row 2: it is [1,2,2], max rect in this histogram is 4; 

For row 3: it is [2,0,3], max rect in this histogram is 3;

Thus we first need to transform the matrix into histograms based on each row(step1), then we use #84 to find out the max rect in the histograms. 

Code:

class Solution {public:    int maximalRectangle(vector<vector<char>>& matrix) {        if(matrix.empty()) return 0;        int m = matrix.size();        int n = matrix[0].size();        vector<vector<int>> hist(m, vector<int>(n,0));<span style="white-space:pre"></span>// step 1        for(int i = 0; i < m; i ++){            for(int j = 0; j < n; j ++){                if(matrix[i][j] == '1'){                    if(i-1>=0) hist[i][j] = hist[i-1][j] + 1;                    else hist[i][j] = 1;                }            }        }<span style="white-space:pre"></span>        int max_area = 0;        for(int i = 0; i < m; i ++){            max_area = max(max_area,largestRectangleArea(hist[i]));        }        return max_area;    }    int largestRectangleArea(vector<int>& heights) {        heights.push_back(0);        stack<int> left_ind;        int max_area = 0;        int i = 0;        while(i < heights.size()){            if(left_ind.empty()) {                left_ind.push(i);                i++;            }            else{                if(heights[i] > heights[left_ind.top()]){                    left_ind.push(i);                    i++;                }else{                    int left = left_ind.top();                    left_ind.pop();                    max_area = max(max_area, heights[left] *(left_ind.empty()?  i  :(i - left_ind.top() - 1)));                }            }                                }        return max_area;     }                    };