剑指 Offer 29. 顺时针打印矩阵

原题链接;剑指 Offer 29. 顺时针打印矩阵

 

solution:

        和螺旋矩阵一样的解法

class Solution {
public:
    int dx[4] = {0, 1, 0, -1},dy[4] = {1, 0, -1, 0};
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector<int> res;    //定义返回值
        if(matrix.size() == 0) return res;
        int m = matrix.size(),n = matrix[0].size();
        vector<vector<bool>> sta(m, vector<bool> (n, false));
        for(int i = 0,x = 0,y = 0,d = 0;i < m * n;i++) {
            res.push_back(matrix[x][y]);
            sta[x][y] = true;

            int a = x + dx[d],b = y + dy[d];
            if(a < 0 || a == m || b < 0 || b == n || sta[a][b]){
                d = (d + 1) % 4;
                a = x + dx[d],b = y + dy[d];
            }
            x = a,y = b;
        }
        return res;
    }
};