[LeetCode]Maximal Rectangle

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.

这题的思路完全来源Largest Rectangle in Histogram，不过要对原始数据做一个处理，把每行处理成直方图形式的数据。然后调用直方图的算法就可以了，所以你的先明白直方图最大面积计算方法。时间复杂度O(n^2)。

输入的矩阵

转换后的矩阵

class Solution {
public:
    int largestRectangleArea(vector<int>& height) {
        int result=0;
        height.push_back(0);
        stack<int> mystack;
        for(int i=0;i<height.size();i++)
        {
            if(mystack.empty() || height[mystack.top()]<=height[i])
            {
                mystack.push(i);
            }
            else
            {
                int index = mystack.top();
                mystack.pop();
                int area = height[index]*(mystack.empty()?i:(i-mystack.top()-1));
                result = max(result,area);
                i--;
            }
        }
        return result;
    }
    
    int maximalRectangle(vector<vector<char>>& matrix) {
        if(matrix.size()==0 || matrix[0].size()==0) return 0;
        int result=0;
        vector<vector<int>> histogram;
        int rows = matrix.size();
        int cols = matrix[0].size();
        vector<int> height_row(cols,0);
        for(int i=0;i<rows;i++)
        {
            for(int j=0;j<cols;j++)
            {
                if(matrix[i][j]=='1')
                {
                    height_row[j]+=1;
                }
                else
                {
                    height_row[j]=0;
                }
            }
            histogram.push_back(height_row);
        }
        for(int i=0;i<rows;i++)
        {
            int maxArea = largestRectangleArea(histogram[i]);
            result = maxArea>result?maxArea:result;
        }
        return result;
    }
};

 其实不需要把整个直方图矩阵得到后再计算，得到每行的直方图举证就可以计算了。

class Solution {
public:
    int largestRectangleArea(vector<int>& height) {
        int result=0;
        height.push_back(0);
        stack<int> mystack;
        for(int i=0;i<height.size();i++)
        {
            if(mystack.empty() || height[mystack.top()]<=height[i])
            {
                mystack.push(i);
            }
            else
            {
                int index = mystack.top();
                mystack.pop();
                int area = height[index]*(mystack.empty()?i:(i-mystack.top()-1));
                result = max(result,area);
                i--;
            }
        }
        return result;
    }
    
    int maximalRectangle(vector<vector<char>>& matrix) {
        if(matrix.size()==0 || matrix[0].size()==0) return 0;
        int result=0;
        int rows = matrix.size();
        int cols = matrix[0].size();
        vector<int> height_row(cols,0);
        for(int i=0;i<rows;i++)
        {
            for(int j=0;j<cols;j++)
            {
                if(matrix[i][j]=='1')
                {
                    height_row[j]+=1;
                }
                else
                {
                    height_row[j]=0;
                }
            }
            int maxArea = largestRectangleArea(height_row);
            result = maxArea>result?maxArea:result;
        }
        return result;
    }
};