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> res = new ArrayList<Integer>();

        int m = matrix.length;

        if(m == 0) return res;

        int n = matrix[0].length;

        //循环次数小于m/2和n/2

        for(int i = 0; i < m/2 && i < n/2; i++){

            for(int j = 0; j < n - 1 - i * 2; j++)

                res.add(matrix[i][i+j]);

            for(int j = 0; j < m - 1 - i * 2; j++)

                res.add(matrix[i+j][n-i-1]);

            for(int j = 0; j < n - 1 - i * 2; j++)

                res.add(matrix[m-i-1][n-i-1-j]);

            for(int j = 0; j < m - 1 - i * 2; j++)

                res.add(matrix[m-i-1-j][i]);

        }

        //循环结束后假设行数/列数是奇数，则还剩一行/列

        if(m % 2 != 0 && m <= n){

            for(int j = 0; j < n - (m/2) * 2; j++)

                res.add(matrix[m/2][m/2+j]);

        }

        else if(n % 2 != 0 && m > n){

            for(int j = 0; j < m - (n/2) * 2; j++)

                 res.add(matrix[n/2+j][n/2]);

        }

        return res;

    }

}