(medium)LeetCode 221.Maximal Square

Given a 2D binary matrix filled with 0's and 1's, find the largest square containing all 1's and return its area.

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Return 4.

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

思路借鉴：dynamic programing.  以当前点(x,y) = '1' 为右下角的最大正方形的边长f(x,y) = min( f(x-1,y), f(x,y-1), f(x-1,y-1)) + 1.

代码如下：

public class Solution {
    public int maximalSquare(char[][] matrix) {
        if(matrix==null || matrix.length==0 || matrix[0].length==0)
            return 0;
        int n=matrix.length;
        int m=matrix[0].length;
        int [][]d=new int[n][m];
        int max=0;
        for(int i=0;i<n;i++){
            if(matrix[i][0]=='1'){
                d[i][0]=1;
                max=1;
            }
        }
        for(int j=0;j<m;j++){
            if(matrix[0][j]=='1'){
                d[0][j]=1;
                max=1;
            }
        }
        for(int i=1;i<n;i++){
            for(int j=1;j<m;j++){
                if(matrix[i][j]=='0') d[i][j]=0;
                else{
                    d[i][j]=Math.min(Math.min(d[i-1][j],d[i][j-1]),d[i-1][j-1])+1;
                    max=Math.max(max,d[i][j]);
                }
            }
        }
        return max*max;

    }
}

　　运行结果：