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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;}