剑指 Offer 29：顺时针打印矩阵

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

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]

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        if(matrix.empty()) return {};
        int m=matrix.size();
        int n=matrix[0].size();
        int left=0,top=0,right=n-1,down=m-1;
        vector<vector<int>> dirs={{0,1},{1,0},{0,-1},{-1,0}};
        int cur_d=0;
        int x=0,y=0;
        vector<int> res;
        for(int i=0;i<m*n;i++){
            int val=matrix[x][y];
            res.push_back(val);
            if(cur_d==0&&y==right){
                top++;
                cur_d++;
            }
            else if(cur_d==1&&x==down){
                right--;
                cur_d++;
            }
            else if(cur_d==2&&y==left){
                down--;
                cur_d++;
            }
            else if(cur_d==3&&x==top){
                left++;
                cur_d++;
            }
            cur_d%=4;
            x+=dirs[cur_d][0];
            y+=dirs[cur_d][1];
        }
        return res;
    }
};