Spiral Matrix

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].

Solution

public class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> result = new ArrayList<>();
        int b = matrix.length;
        if (b == 0) {
            return new ArrayList<>();
        }

        int a = matrix[0].length;

        if (a == 0 && b == 0) {
            return new ArrayList<>();
        } else if (a == 1) {
            for (int i = 0; i < b; i++) {
                result.add(matrix[i][0]);
            }
            return result;
        } else if (b == 1) {
            for (int i = 0; i < a; i++) {
                result.add(matrix[0][i]);
            }

            return result;
        }

        spiral(matrix, b, a, 0, result);

        return result;
    }

    public void spiral(int[][] matrx, int b, int a, int k, List<Integer> result) {
        if (k > a - k - 1 || k > b - k - 1) {
            return;
        }

        if (k == a - k - 1 && k == b - k - 1) {
            result.add(matrx[k][k]);
            return;
        }

        for (int i = k; i < a - k - 1; i++) {
            result.add(matrx[k][i]);
        }


        for (int i = k; i < b - k - 1; i++) {
            result.add(matrx[i][a - k - 1]);
        }

        if (k == b - k - 1) {
            result.add(matrx[b - k - 1][a - k - 1]);
        } else {

            for (int i = a - k - 1; i > k; i--) {
                result.add(matrx[b - k - 1][i]);
            }
        }
        if (k == a - k - 1) {
            result.add(matrx[b - k - 1][k]);
        } else {
            for (int i = b - k - 1; i > k; i--) {
                result.add(matrx[i][k]);
            }
        }


        spiral(matrx, b, a, k + 1, result);
    }
}