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剑指 Offer 29. 顺时针打印矩阵
2022-07-06 02:11:00 【anieoo】
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;
}
};
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