Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
For example,Given the following matrix:
[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]
You should return [1,2,3,6,9,8,7,4,5].

public class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> rec = new ArrayList<Integer>();
        
        if(matrix.length == 0||matrix[0].length == 0||matrix == null)
           return rec;
        int m = matrix.length,n = matrix[0].length;
        int x1 = 0,y1 = 0,x2=m-1,y2=n-1;
        while(m>1&&n>1)
        {
            x2 = x1+m-1;
            y2 = y1+n-1;
            
            for(int j=y1;j<y2;j++)
            {
                rec.add(matrix[x1][j]);
                System.out.println(matrix[x1][j]);
            }
            for(int i=x1;i<x2;i++)
            {
                rec.add(matrix[i][y2]);
                System.out.println(matrix[i][y2]);
            }
            
            for(int j=y2;j>y1;j--)
            {
                rec.add(matrix[x2][j]);
                System.out.println(matrix[x2][j]);
            }
          
            for(int i=x2;i>x1;i--)
            {
                rec.add(matrix[i][y1]);
                System.out.println(matrix[i][y1]);
            }
          
            x1++;
            y1++;
            m -= 2;
            n -= 2;
            
        }
        x2 = x1+m-1;
        y2 = y1+n-1;
        if(m==1&&n!=0)
        {
           for(int i=y1;i<=y2;i++)
              rec.add(matrix[x2][i]);
           return rec;
        }
        if(n==1&&m!=0)
        {
           for(int i=x1;i<=x2;i++)
              rec.add(matrix[i][y2]);
           return rec;
        }
        return rec;
    }
}