将一个二维矩阵，螺旋方式遍历。
Example 1:
Input: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] Output: [1,2,3,6,9,8,7,4,5]Example 2:
Input: [ [1, 2, 3, 4], [5, 6, 7, 8], [9,10,11,12] ] Output: [1,2,3,4,8,12,11,10,9,5,6,7]思考：本题的难点在于以螺旋的方式访问。需要考虑各个方向和是否访问过，何时旋转方向继续向前。
解法：
方法1：使用辅助空间记录是否已经访问过。
vector<int> spiralOrder(vector<vector<int>>& matrix) { vector<int> res; if (matrix.empty()) { return res;} int rows = matrix.size(); int cols = matrix[0].size(); vector<vector<bool>> visited(rows, vector<bool>(cols, false)); int len = rows * cols; int r = 0, c = 0; int j = 0; int dr[4] = {0, 1, 0, -1}; int dc[4] = {1, 0, -1, 0}; for (int i = 0; i < len; ++i) { res.push_back(matrix[r][c]); visited[r][c] = true; int n_r = r + dr[j]; int n_c = c + dc[j]; if ((n_r >= 0 && n_r < rows) && (n_c >= 0 && n_c < cols) && (!visited[n_r][n_c])) { r = n_r; c = n_c; } else { j = (j + 1) % 4; r += dr[j]; c += dc[j]; } } return res; }方法2：直接遍历。需要注意while，for循环的比较运算符。
vector<int> spiralOrder(vector<vector<int>>& matrix) { vector<int> res; if (matrix.empty() || matrix[0].empty()) { return res; } int r1 = 0; int r2 = matrix.size() - 1; int c1 = 0; int c2 = matrix[0].size() - 1; while(r1 <= r2 && c1 <= c2) { for (int c = c1; c <= c2; ++c) { res.push_back(matrix[r1][c]); } for (int r = r1 + 1; r <= r2; ++r) { res.push_back(matrix[r][c2]); } if (r1 < r2 && c1 < c2) { for (int c = c2 - 1; c > c1; --c) { res.push_back(matrix[r2][c]); } for (int r = r2; r > r1; --r) { res.push_back(matrix[r][c1]); } } r1++; r2--; c1++; c2--; } return res; }
