剑指 Offer 29. 顺时针打印矩阵 && Leetcode 54. 螺旋矩阵

输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字。

示例 1：
输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出：[1,2,3,6,9,8,7,4,5]

示例 2：
输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
输出：[1,2,3,4,8,12,11,10,9,5,6,7]
 

限制：
0 <= matrix.length <= 100
0 <= matrix[i].length <= 100
注意：本题与主站 54 题相同：https://leetcode-cn.com/problems/spiral-matrix/

    int addx[4] = {0,1,0,-1};
    int addy[4] = {1,0,-1,0};

class Solution {
public:
    int addx[4] = {0,1,0,-1};
    int addy[4] = {1,0,-1,0};
    vector<vector<int>> vis;
    vector<int> ans;
    void dfs(int x,int y, int idx,const vector<vector<int>>& matrix){
        if(vis[x][y] ==0){
            vis[x][y] = 1;
            ans.push_back(matrix[x][y]);
        }

        for(int i = 0; i < 4;i++){
            int newx = x+addx[(idx+i)%4];
            int newy = y+addy[(idx+i)%4];
            if(newx>=0 && newx<matrix.size() && newy>=0 && newy<matrix[0].size() && 
                vis[newx][newy] == 0 )
            {
                dfs(newx,newy,(idx+i)%4,matrix);
                break;
            }
        }
        return;
    }

    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        if(matrix.empty()) return ans;
        int n = matrix.size(); int m  = matrix[0].size();
        vis= vector<vector<int>>(n,vector<int>(m));
        dfs(0,0,0,matrix);
        return ans;
    }
};