leetcode 54. 螺旋矩陣

給定一個包含 m x n 個元素的矩陣（m 行, n 列），請按照順時針螺旋順序，返回矩陣中的所有元素。

示例 1:

輸入:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
輸出: [1,2,3,6,9,8,7,4,5]
示例 2:

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

直接用模擬法即可。定義上下左右四個邊界，每向一個方向遍歷一次後，將邊界移動，然後判斷邊界是否合法，不合法說明遍歷完了，可直接返回list

 public List<Integer> spiralOrder(int[][] matrix) {
         if (matrix.length == 0 || matrix[0].length == 0) {
            return new ArrayList<>();
        }
        List<Integer> list = new ArrayList<>();
        int rowUp = 0;
        int rowDown = matrix.length - 1;
        int colLeft = 0;
        int colRight = matrix[0].length - 1;
        int direction = 0;
        while (direction != -1) {
            if (direction == 0) {
                //往右
                for (int i = colLeft; i <= colRight; i++) {
                    list.add(matrix[rowUp][i]);
                }
                rowUp++;
                if (rowUp > rowDown) {
                    return list;
                }
                direction = 1;
            } else if (direction == 1) {
                //往下
                for (int i = rowUp; i <= rowDown; i++) {
                    list.add(matrix[i][colRight]);
                }
                colRight--;
                if (colRight < colLeft) {
                    return list;
                }
                direction = 2;
            } else if (direction == 2) {
                //往左
                for (int i = colRight; i >= colLeft; i--) {
                    list.add(matrix[rowDown][i]);
                }
                rowDown--;
                if (rowDown < rowUp) {
                    return list;
                }
                direction = 3;
            } else if (direction == 3) {
                //往上
                for (int i = rowDown; i >= rowUp; i--) {
                    list.add(matrix[i][colLeft]);
                }
                colLeft++;
                if (colLeft > colRight) {
                    return list;
                }
                direction = 0;
            }
        }
        return list;
    }