解題思路：
\qquad 這道題可以直接用模擬解決，順時(shí)針螺旋可以分解為依次沿“右-下-左-上”四個(gè)方向的移動(dòng)，每次碰到“邊界”時(shí)改變方向，邊界是不可到達(dá)或已經(jīng)到達(dá)過(guò)的地方，會(huì)隨著指針移動(dòng)不斷收縮。

	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;}