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.

Analysis:

For each position (i,j), we calculate how many accumlative 1s at the left direction of this position at this row. We record this number at d[i][j]. For example, for matrix:

0 0 0 1 1 1

0 1 1 1 1 1

0 1 1 1 1 1

1 0 1 1 1 1

We have matrix d[i][j]

0 0 0 1 2 3

0 1 2 3 4 5

0 1 2 3 4 5

1 0 1 2 3 4

Then, at each postion (i,j), we scan the matrix along the upper direction and calculate the area of the maximum 1 matrix with the right bottom corner at (i,j). For row k<i, the 1s matrix should be (k-i+1)*min(d[l][j], l=k...i);

After caluclating the 1s matrix for each (i,j), we get the answer.

Solution:

public class Solution {
    public int maximalRectangle(char[][] matrix) {
        int rowNum = matrix.length;
        if (rowNum==0) return 0;
        int colNum = matrix[0].length;
        if (colNum==0) return 0;

        int[][] d = new int[rowNum][colNum];
        for (int i=0;i<rowNum;i++)
            for (int j=0;j<colNum;j++)
                if (j==0)
                    if (matrix[i][j]=='1') d[i][j]=1;
                    else d[i][j]=0;
                else
                    if (matrix[i][j]=='1') d[i][j]=d[i][j-1]+1;
                    else d[i][j]=0;

        int maxSoFar = -1;
        for (int i=0;i<rowNum;i++)
            for (int j=0;j<colNum;j++){
                int curMax = d[i][j];
                int minLen = d[i][j];
                for (int k=(i-1);k>=0;k--){
                    if (d[k][j]==0) break;
                    if (d[k][j]<minLen) minLen = d[k][j];
                    int newMax = (i-k+1)*minLen;
                    if (curMax<newMax) curMax = newMax;
                }
                if (curMax>maxSoFar) maxSoFar = curMax;

            }

        return maxSoFar;
    }
}