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].
Array
思路：dfs
由于边界条件太多，直接构造新的数组，其边缘都有INT_MIN填充，方便边界判断
以向右为例，能向右就向右，不能向右则向下。。。 以此类推
class Solution { enum direct{RIGHT = 0, DOWN, LEFT, UP}; vector<int> m_res; public: #if 1 void dfs(int i, int j, enum direct d, vector<vector<int> > &matrix) { if(matrix[i][j] != INT_MIN)// this juge can be deleted { //cout << i <<",\t" << j << endl; m_res.push_back(matrix[i][j]); matrix[i][j] = INT_MIN; if(d == RIGHT) { if(matrix[i][j+1] != INT_MIN) dfs(i, j+1, RIGHT, matrix); else if(matrix[i+1][j] != INT_MIN) dfs(i+1, j, DOWN, matrix); } else if(d == DOWN) { if(matrix[i+1][j] != INT_MIN) dfs(i+1, j, DOWN, matrix); else if(matrix[i][j-1] != INT_MIN) dfs(i, j-1, LEFT, matrix); } else if(d == LEFT) { if(matrix[i][j-1] != INT_MIN) dfs(i, j-1, LEFT, matrix); else if(matrix[i-1][j] != INT_MIN) dfs(i-1, j, UP, matrix); } else if(d == UP) { if(matrix[i-1][j] != INT_MIN) dfs(i-1, j, UP, matrix); else if(matrix[i][j+1] != INT_MIN) dfs(i, j+1, RIGHT, matrix); } } } #endif vector<int> spiralOrder(vector<vector<int> > &matrix) { if(matrix.size() == 0 || matrix[0].size() == 0) return m_res; int row = matrix.size(); int col = matrix[0].size(); vector<vector<int> > newMat; vector<int> intVec(col+2, INT_MIN); newMat.push_back(intVec); for(int i = 0; i < row; i++) { for(int j = 0; j < col; j++) { intVec[j+1] = matrix[i][j]; } newMat.push_back(intVec); } intVec.clear(); intVec.resize(col+2, INT_MIN); newMat.push_back(intVec); //for(int i = 0; i < newMat.size(); i++) // printVector(newMat[i]); dfs(1,1,RIGHT,newMat); return m_res; } };
原文：http://www.cnblogs.com/diegodu/p/4314079.html