[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.

DP。用f[i][j]來記錄i行以j列為結尾，往前連續的1的個數。然后再一個O(n^3)的循環來找以(i, j)為右下角的矩形最大的1的面積。

class Solution {
private:
    int f[1000][1000];
public:
    int maximalRectangle(vector<vector<char> > &matrix) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        for(int i = 0; i < matrix.size(); i++)
            f[i][0] = matrix[i][0] == '1' ? 1 : 0;
            
        for(int i = 0; i < matrix.size(); i++)
            for(int j = 1; j < matrix[i].size(); j++)
                f[i][j] = matrix[i][j] == '1' ? f[i][j-1] + 1 : 0;
                
        int ret = 0;
        
        for(int i = 0; i < matrix.size(); i++)
            for(int j = 0; j < matrix[i].size(); j++)
            {
                int k = i;
                int width = INT_MAX;
                
                while(k >= 0)
                {
                    if (f[k][j] == 0)
                        break;
                    
                    width = min(width, f[k][j]);
                
                    ret = max(ret, width * (i - k + 1));   
                    
                    k--;                   
                }
            }
            
        return ret;
    }
};

 

 O(n^2)的算法，可以把問題看成求多個直方圖的最大矩形面積，這樣就可以從上往下來做例如第i層就是0~i的直方圖，第i+1就是0~i+1的直方圖，這樣求直方圖的復雜度O(n)，於是就有O(n^2)

class Solution {
public:
    int calArea(vector<int> &a, vector<int> &width)
    {
        for(int i = 0; i < width.size(); i++)
            width[i] = 0;
            
        stack<int> s;
        for(int i = 0; i < a.size(); i++)
            if (s.empty())
            {
                s.push(i);
                width[i] = 0;
            }
            else
            {
                while(!s.empty())
                {
                    if (a[s.top()] < a[i])
                    {
                        width[i] = i - s.top() - 1;
                        s.push(i);
                        break;
                    }
                    else
                        s.pop();
                }
                
                if (s.empty())
                {
                    s.push(i);
                    width[i] = i;
                }
            }
            
        while(!s.empty())
            s.pop();
            
        for(int i = a.size() - 1; i >= 0; i--)
            if (s.empty())
            {
                s.push(i);
                width[i] = 0;
            }
            else
            {
                while(!s.empty())
                {
                    if (a[i] > a[s.top()])
                    {
                        width[i] += s.top() - i - 1;
                        s.push(i);
                        break;
                    }
                    else
                        s.pop();
                }
                
                if (s.empty())
                {
                    width[i] += a.size() - i - 1;
                    s.push(i);
                }
            }
        
        int maxArea = 0;    
        for(int i = 0; i < width.size(); i++)
            maxArea = max(maxArea, (width[i] + 1) * a[i]);
            
        return maxArea;
    }
    
    int maximalRectangle(vector<vector<char> > &matrix) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (matrix.size() == 0)
            return 0;
            
        vector<int> a(matrix[0].size(), 0);
        vector<int> width(matrix[0].size());
        
        int maxArea = 0;
        for(int i = 0; i < matrix.size(); i++)
        {
            for(int j = 0; j < matrix[i].size(); j++)
                a[j] = matrix[i][j] == '1' ? a[j] + 1 : 0;
                
            maxArea = max(maxArea, calArea(a, width));
        }
        
        return maxArea;
    }
};