给定一个包含 m x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。
示例 1:
输入:
[
[ 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:
[
[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]
代码
class Solution {
public:
vector<int> spiralOrder(vector<vector<int> > &matrix) {
vector<int> res;
if (matrix.empty() || matrix[0].empty()) return res;
int m = matrix.size(), n = matrix[0].size();
int c = m > n ? (n + 1) / 2 : (m + 1) / 2;
int p = m, q = n;
for (int i = 0; i < c; ++i, p -= 2, q -= 2) {
for (int col = i; col < i + q; ++col)
res.push_back(matrix[i][col]);
for (int row = i + 1; row < i + p; ++row)
res.push_back(matrix[row][i + q - 1]);
if (p == 1 || q == 1) break;
for (int col = i + q - 2; col >= i; --col)
res.push_back(matrix[i + p - 1][col]);
for (int row = i + p - 2; row > i; --row)
res.push_back(matrix[row][i]);
}
return res;
}
};
