每次横向（纵向）走到尽头调换方向，下次横向（纵向）经过的元素便减一个。
时间复杂度:O(M*N)
C++代码：

class Solution {
public:
	enum dir { UP, DOWN, LEFT, RIGHT };
	vector<int> spiralOrder(vector<vector<int>>& matrix) {
		if (matrix.empty() || matrix[0].empty())
			return {};
		if (matrix.size() == 1)
			return matrix[0];
		vector<int> result;
		for (auto x : matrix[0])
			result.push_back(x);
		int x = matrix[0].size() - 1, y = 0;
		int col = matrix.size() - 1;
		int row = matrix[0].size() - 1;
		dir direct = DOWN;
		while (col || row)
		{
			switch (direct)
			{
			case RIGHT:
			{
				for (int i = 0; i < row; i++)
					result.push_back(matrix[y][++x]);
				if (col == 0)
					row = 1;
				row--;
				direct = DOWN;
				break;
			}
			case DOWN:
			{
				for (int i = 0; i < col; i++)
					result.push_back(matrix[++y][x]);
				if (row == 0)
					col = 1;
				col--;
				direct = LEFT;
				break;
			}
			case LEFT:
			{
				for (int i = 0; i < row; i++)
					result.push_back(matrix[y][--x]);
				if (col == 0)
					row = 1;
				row--;
				direct = UP;
				break;
			}
			case UP:
			{
				for (int i = 0; i < col; i++)
					result.push_back(matrix[--y][x]);
				if (row == 0)
					col = 1;
				col--;
				direct = RIGHT;
				break;
			}
			}
		} 
		return result;
	}
};