标签:ref solution tps top amp 空间复杂度 判断 空间 for
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]
依次从四个方向判断。
时间复杂度:O(n*m)
空间复杂度:O(1)
class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
vector<int> res;
if (matrix.empty()) return res;
int row = matrix.size(), col = matrix[0].size();
int top = 0, bottom = row - 1, left = 0, right = col - 1;
while (true) {
for (int j = left; j <= right; ++j) res.push_back(matrix[top][j]);
if (++top > bottom) break;
for (int i = top; i <= bottom; ++i) res.push_back(matrix[i][right]);
if (--right < left) break;
for (int j = right; j >= left; --j) res.push_back(matrix[bottom][j]);
if (--bottom < top) break;
for (int i = bottom; i >= top; --i) res.push_back(matrix[i][left]);
if (++left > right) break;
}
return res;
}
};
标签:ref solution tps top amp 空间复杂度 判断 空间 for
原文地址:https://www.cnblogs.com/galaxy-hao/p/12709241.html