[LeetCode]--54. Spiral Matrix

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].

我想着把这个二维数组分成一层一层的环，然后一层一层的遍历，就能得到，而遍历环就是上边右边下边左边。就是要控制好循环结束之类的。

public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> res = new ArrayList<Integer>();
        if (matrix == null || matrix.length == 0)
            return res;
        int size = matrix.length * matrix[0].length;
        int len = Math.min(matrix.length, matrix[0].length);
        for (int i = 0; i < len; i++) {
            if (res.size() == size)
                break;
            visitCircle(res, matrix, i);
        }
        return res;
    }

    public void visitCircle(List<Integer> list, int[][] a, int m) {
        int len = a[0].length, hlen = a.length;
        for (int i = m; i < len - m; i++) {
            list.add(a[m][i]);
        }
        for (int i = 1 + m; i < hlen - m; i++) {
            list.add(a[i][len - m - 1]);
        }
        if (hlen - 2 * m == 1 || len - 2 * m == 1)
            return;
        for (int i = len - 2 - m; i >= m; i--) {
            list.add(a[hlen - m - 1][i]);
        }
        for (int i = hlen - m - 2; i > m; i--) {
            list.add(a[i][m]);
        }
    }

还有一种方法，跟我这个其实是一个意思，但是他没有这样表达。

public ArrayList<Integer> spiralOrder(int[][] matrix) {
        ArrayList<Integer> rst = new ArrayList<Integer>();
        if (matrix == null || matrix.length == 0)
            return rst;

        int rows = matrix.length;
        int cols = matrix[0].length;
        int count = 0;
        while (count * 2 < rows && count * 2 < cols) {
            for (int i = count; i < cols - count; i++)
                rst.add(matrix[count][i]);

            for (int i = count + 1; i < rows - count; i++)
                rst.add(matrix[i][cols - count - 1]);

            if (rows - 2 * count == 1 || cols - 2 * count == 1)
                // if only one row /col remains
                break;

            for (int i = cols - count - 2; i >= count; i--)
                rst.add(matrix[rows - count - 1][i]);

            for (int i = rows - count - 2; i >= count + 1; i--)
                rst.add(matrix[i][count]);

            count++;
        }
        return rst;
    }