Q54 螺旋矩阵

给定一个包含 m x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。

示例 1:

输入:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
输出: [1,2,3,6,9,8,7,4,5]

示例 2:

输入:
[
  [1, 2, 3, 4],
  [5, 6, 7, 8],
  [9,10,11,12]
]
输出: [1,2,3,4,8,12,11,10,9,5,6,7]

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> list = new ArrayList<>();

        if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
            return list;

        int lr = 0;
        int lc = 0;
        int rr = matrix.length - 1;
        int rc = matrix[0].length - 1;
        while (lr <= rr && lc <= rc)
            print(matrix, lr++, lc++, rr--, rc--, list);

        return list;
    }

    private void print(int[][] matrix, int lr, int lc, int rr, int rc, List<Integer> list) {
        if (lr == rr) {
            for (int i = lc; i <= rc; i++)
                list.add(matrix[lr][i]);
        } else if (lc == rc) {
            for (int i = lr; i <= rr; i++)
                list.add(matrix[i][lc]);
        } else {
            for (int i = lc; i < rc; i++)
                list.add(matrix[lr][i]);

            for (int i = lr; i < rr; i++)
                list.add(matrix[i][rc]);

            for (int i = rc; i > lc; i--)
                list.add(matrix[rr][i]);

            for (int i = rr; i > lr; i--)
                list.add(matrix[i][lc]);
        }

    }
}