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].
典型的dfs问题，与前面一个问题比较像，代码如下。单独维护一个数组记录是否已经走过某个格子，注意边界条件的判断就可以了，代码如下：
class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
drt = vector<vector<int>>{{,},{,},{-,},{,-}}; //四个方向
mark = vector<vector<bool>>(,vector<bool>(,false));
ret.clear();
if(!matrix.size() || !matrix[].size())
return ret;
dfs(matrix, , , -);//这里取-1是为了能从(0,0)开始。
return ret;
}
void dfs(vector<vector<int>> & matrix, int direct, int x, int y)
{
for(int i = ; i < ; ++i){
int j = (direct + i) % ;
int tx = x + drt[j][];
int ty = y + drt[j][];
if(tx >= && tx < matrix.size() &&
ty >= && ty < matrix[].size() &&
mark[tx][ty] == false){
mark[tx][ty] = true;
ret.push_back(matrix[tx][ty]);
dfs(matrix, j, tx, ty);
}
}
}
private:
vector<vector<int>> drt;
vector<vector<bool>> mark;
vector<int> ret;
};