class Solution { public: vector<int> spiralOrder(vector<vector<int> > &matrix) { vector<int> ret; int cols = 0; int rows = 0; if (!(rows = matrix.size()) || !(cols = matrix[0].size())) { return ret; } int left = 0, top = 0, right = cols - 1, bottom = rows - 1; while (left <= right && top <= bottom) { circle_walk(left, top, right, bottom, matrix, ret); left++, top++, bottom--, right--; } return ret; } void circle_walk(int left, int top, int right, int bottom, vector<vector<int> >& map, vector<int>& path) { int rows = 0; if (map.size() < 1 || map[0].size() < 1) return; // top row for (int i=left; i<=right; i++) { path.push_back(map[top][i]); } // right col for (int i=top+1; i<=bottom; i++) { path.push_back(map[i][right]); } // bottom row for (int i=right-1; bottom != top && i>=left; i--) { path.push_back(map[bottom][i]); } // left col for (int i=bottom-1; left != right && i>top; i--) { path.push_back(map[i][left]); } } };
又隔了好久没有写了,水一发,准备测试数据还是必要的,比如只有一行的matrix,只有一列的matrix等,或许可以发现编码中的问题。
第二轮:
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
For example,
Given the following matrix:
[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]
You should return [1,2,3,6,9,8,7,4,5]
.
题解简洁很多:
class Solution { public: vector<int> spiralOrder(vector<vector<int> > &matrix) { vector<int> res; int m = matrix.size(); if (m == 0) { return res; } int n = matrix[0].size(); int row = 0, col = -1; while(true) { for (int i=0; i<n; i++) res.push_back(matrix[row][++col]); if (--m == 0) break; for (int i=0; i<m; i++) res.push_back(matrix[++row][col]); if (--n == 0) break; for (int i=0; i<n; i++) res.push_back(matrix[row][--col]); if (--m == 0) break; for (int i=0; i<m; i++) res.push_back(matrix[--row][col]); if (--n == 0) break; } return res; } };