螺旋矩阵

给你一个 m 行 n 列的矩阵 matrix ，请按照 顺时针螺旋顺序 ，返回矩阵中的所有元素。

示例 1：

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

示例 2：

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输入：matrix = [[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]

提示：

  • m == matrix.length
  • n == matrix[i].length
  • 1 <= m, n <= 10
  • -100 <= matrix[i][j] <= 100

定义两个数组模拟移动方向

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class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        int[] dx = new int[] { 0, 1, 0, -1 };
        int[] dy = new int[] { 1, 0, -1, 0 };
        List<Integer> list = new ArrayList<>();
        int d = 0;
        int x = 0;
        int y = 0;
        int m = matrix.length;
        int n = matrix[0].length;
        for (int i = 0; i < m * n; i++) {
            list.add(matrix[x][y]);
            matrix[x][y] = 200;
            int nextX = x + dx[d];
            int nextY = y + dy[d];
            if (nextX < 0 || nextY < 0 || nextX >= m || nextY >= n || matrix[nextX][nextY] == 200) {
                d = (d + 1) % 4;
                nextX = x + dx[d];
                nextY = y + dy[d];
            }
            x = nextX;
            y = nextY;
        }
        return list;
    }
}