Question:

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

Example 1:

Input:
[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]

Example 2:

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

Thinking:

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> result = new ArrayList<>();
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return result;
        int left = 0; int right = matrix[0].length - 1;
        int up = 0; int down = matrix.length - 1;
        int dir = 0; // 0:right, 1:down, 2:left, 3:up
        int tempx = 0; int tempy = 0;
        result.add(matrix[0][0]);
        while(down >= up && right >= left){
            switch(dir){
                case 0:
                    while(++tempy <= right){
                        result.add(matrix[tempx][tempy]);
                    }
                    tempy--;
                    up++;
                    dir = 1;
                    break;
                case 1:
                    while(++tempx <= down){
                        result.add(matrix[tempx][tempy]);
                    }
                    tempx--;
                    right--;
                    dir = 2;
                    break;
                case 2:
                    while(--tempy >= left){
                        result.add(matrix[tempx][tempy]);
                    }
                    tempy++;
                    down--;
                    dir = 3;
                    break;
                case 3:
                    while(--tempx >= up){
                        result.add(matrix[tempx][tempy]);
                    }
                    tempx++;
                    left++;
                    dir = 0;
                    break;
                default:
                    break;
            }
        }
        return result;
    }
}

二刷

还是用了一刷的办法，使用case-switch来控制循环。

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> result = new LinkedList<>();
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return result;
        int height = matrix.length, width = matrix[0].length;
        int up = 0, down = height - 1, left = 0, right = width - 1;
        int dir = 0; //0:right, 1:down, 2:left, 3: up
        int x = 0, y = 0;
        result.add(matrix[0][0]);
        while(up <= down && right >= left){
            switch(dir){
                case 0:
                    while(++y <= right)
                        result.add(matrix[x][y]);
                    up++; y--;
                    dir = 1;
                    break;
                case 1:
                    while(++x <= down)
                        result.add(matrix[x][y]);
                    right--; x--;
                    dir = 2;
                    break;
                case 2:
                    while(--y >= left)
                        result.add(matrix[x][y]);
                    down--; y++;
                    dir = 3;
                    break;
                case 3:
                    while(--x >= up)
                        result.add(matrix[x][y]);
                    left++; x++;
                    dir = 0;
                    break;
                default:
                    break;
            }
        }
        return result;
    }
}