Question:

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].

Thinking:

There are four directions can go, move according to the order of right, down, left, up and check the bound of moving.

Solution:

public List<Integer> spiralOrder(int[][] matrix) {
    List<Integer> res = new ArrayList<Integer>();
    int m = matrix.length;
    if (m == 0)    return res;
    int n = matrix[0].length;
    int direction = 0;//0right, 1down, 2left, 3up
    int i = 0, j = 0;//position
    int c0 = 1, c1 = 1, c2 = 0, c3 = 1;//row and column limit

    while (c0 + c2 - 1 <= n && c1 + c3 - 1 <= m) {
        res.add(matrix[i][j]);
        switch (direction) {
        case 0:
            if (j == n-c0){
                direction = (direction+1) % 4;
                c0++;
                res.remove(res.size()-1);
            }
            else
                j++;
            break;
        case 1:
            if (i == m-c1){
                direction = (direction+1) % 4;
                c1++;
                res.remove(res.size()-1);
            }
            else
                i++;
            break;
        case 2:
            if (j == c2){
                direction = (direction+1) % 4;
                c2++;
                res.remove(res.size()-1);
            }
            else
                j--;
            break;
        case 3:
            if (i == c3){
                direction = (direction+1) % 4;
                c3++;
                res.remove(res.size()-1);
            }
            else
                i--;
            break;
        }
    }
    res.add(matrix[i][j]);

    return res;
}