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.

O(n^3)方法，348 milli secs

 

class Solution {
public:
    int maximalRectangle(vector<vector<char> > &matrix) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int y = matrix.size();
        if (y == 0) return 0;
        int x = matrix[0].size();
        if (x == 0) return 0;
        
        vector< vector<int> > result(y, vector<int>(x));
        for (int j = 0; j < y; ++j)
        {
            for (int i = 0; i < x; ++i)
            {
                result[j][i] = '0' == matrix[j][i] ? 0 : 1;
            }
        }
        
        for (int j = 0; j < y; ++j)
        {
            for (int i = 1; i < x; ++i)
            {
                result[j][i] += (0 == result[j][i] ? 0 : result[j][i-1]);
            }
        }
        
        int max = 0;
        for (int i = 0; i < x; ++i)
        {
            for (int j = 0; j < y; ++j)
            {
                int maxWidth = result[j][i];  
                for (int k = j; k < y; ++k)
                {
                    maxWidth = maxWidth <  result[k][i] ? maxWidth : result[k][i];
                    max = max > maxWidth*(k-j+1) ? max : maxWidth*(k-j+1);
                }
            }
        }
        return max;
    }
};

O(n^2)方法： 44 milli secs

class Solution {
public:
    int maximalRectangle(vector<vector<char> > &matrix) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int y = matrix.size();
        if (y == 0) return 0;
        int x = matrix[0].size();
        if (x == 0) return 0;
        
        vector< vector<int> > result(y, vector<int>(x));
        for (int j = 0; j < y; ++j)
        {
            int tmp = 0;
            for (int i = 0; i < x; ++i)
            {
                if ('0' == matrix[j][i])
                    tmp =  0;
                else
                    ++tmp;
                result[j][i] = tmp;
            }
        }
        
        int max = 0;
        int* l = new int[y];  
        int* r = new int[y];  
        for (int i = 0; i < x; ++i)
        {
            l[0] = 0;
            for (int j = 1; j < y; ++j)
            {
                int k = j;  
                while(k > 0 && result[j][i] <= result[k-1][i])  
                    k = l[k-1];  
                l[j] = k;  
            }
            
            r[y-1] = y - 1; 
            for(int j = y-2; j >= 0; --j)  
            {  
                int k = j;  
                while(k < y - 1 && result[j][i] <= result[k+1][i])  
                    k = r[k+1];  
                r[j] = k;  
            } 
            
            for (int j = 0; j < y; ++j)  
            {  
                int ar = (r[j] - l[j] + 1) * result[j][i];  
                max = max > ar ? max : ar;     
            }  
        }

        return max;
    }
};