解题思路：
\qquad 这道题可以直接用模拟解决，顺时针螺旋可以分解为依次沿“右-下-左-上”四个方向的移动，每次碰到“边界”时改变方向，边界是不可到达或已经到达过的地方，会随着指针移动不断收缩。

	vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector<int> ans;
        int mode = 0, i = 0, j = 0;
        int a = matrix[0].size(), b = matrix.size();
        vector<int> bound = {a, b, -1, 0};

        for(int m = 0; m < matrix.size() * matrix[0].size(); m++)
        {
            ans.push_back(matrix[i][j]);

            switch(mode)
            {
                case 0:
                    if(j+1 >= bound[0])
                    {
                        bound[0] = j;
                        mode = (mode + 1) % 4;
                        i++;
                    }
                    else j++;
                    break;
                case 1:
                    if(i + 1 >= bound[1])
                    {
                        bound[1] = i;
                        mode = (mode + 1) % 4;
                        j--;
                    }
                    else i++;
                    break;
                case 2:
                    if(j-1 <= bound[2])
                    {
                        bound[2] = j;
                        mode = (mode + 1) % 4;
                        i--;
                    }
                    else j--;
                    break;
                case 3:
                    if(i - 1 <= bound[3])
                    {
                        bound[3] = i;
                        mode = (mode + 1) % 4;
                        j++;
                    }
                    else i--;
                    break;

                default:
                    break;
            }
        }
        return ans;
    }