// 54. Spiral Matrix #include "leetcode.h" // given an m x n 'matrix', return all elements of the 'matrix' in spiral order class Solution { public: vector spiralOrder(vvd & matrix) { vvd dirs{{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; int rows = matrix.size(), cols = matrix[0].size(); vector ans, n_step{cols, rows - 1}; if (!rows || !cols) return ans; int i_dir = 0, i_row = 0, i_col = -1; while (n_step[i_dir % 2]) { for (int i = 0; i < n_step[i_dir % 2]; ++i) { i_row += dirs[i_dir][0]; i_col += dirs[i_dir][1]; ans.push_back(matrix[i_row][i_col]); } n_step[i_dir % 2]--; i_dir = (i_dir + 1) % 4; } return ans; } }; int main() { Solution obj; vvd m1 = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}; vvd m2 = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}}; for (auto i : obj.spiralOrder(m1)) printf("%d, ", i); printf("\n"); // expect: 1,2,3,6,9,8,7,4,5 for (auto i : obj.spiralOrder(m2)) printf("%d, ", i); // expect: 1,2,3,4,8,12,11,10,9,5,6,7 }