• Time:O(mn)
• Space:O(1)

## C++

``````class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
if (matrix.empty())
return {};

const int m = matrix.size();
const int n = matrix[0].size();
vector<int> ans;
int r1 = 0;
int c1 = 0;
int r2 = m - 1;
int c2 = n - 1;

// repeatedly add matrix[r1..r2][c1..c2] to ans
while (ans.size() < m * n) {
for (int j = c1; j <= c2 && ans.size() < m * n; ++j)
ans.push_back(matrix[r1][j]);
for (int i = r1 + 1; i <= r2 - 1 && ans.size() < m * n; ++i)
ans.push_back(matrix[i][c2]);
for (int j = c2; j >= c1 && ans.size() < m * n; --j)
ans.push_back(matrix[r2][j]);
for (int i = r2 - 1; i >= r1 + 1 && ans.size() < m * n; --i)
ans.push_back(matrix[i][c1]);
++r1, ++c1, --r2, --c2;
}

return ans;
}
};
``````

## JAVA

``````class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
if (matrix.length == 0)
return new ArrayList<>();

final int m = matrix.length;
final int n = matrix[0].length;
List<Integer> ans = new ArrayList<>();
int r1 = 0;
int c1 = 0;
int r2 = m - 1;
int c2 = n - 1;

// repeatedly add matrix[r1..r2][c1..c2] to ans
while (ans.size() < m * n) {
for (int j = c1; j <= c2 && ans.size() < m * n; ++j)
ans.add(matrix[r1][j]);
for (int i = r1 + 1; i <= r2 - 1 && ans.size() < m * n; ++i)
ans.add(matrix[i][c2]);
for (int j = c2; j >= c1 && ans.size() < m * n; --j)
ans.add(matrix[r2][j]);
for (int i = r2 - 1; i >= r1 + 1 && ans.size() < m * n; --i)
ans.add(matrix[i][c1]);
++r1;
++c1;
--r2;
--c2;
}

return ans;
}
}
``````

## Python

``````class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
if not matrix:
return []

m = len(matrix)
n = len(matrix[0])
ans = []
r1 = 0
c1 = 0
r2 = m - 1
c2 = n - 1

# repeatedly add matrix[r1..r2][c1..c2] to ans
while len(ans) < m * n:
j = c1
while j <= c2 and len(ans) < m * n:
ans.append(matrix[r1][j])
j += 1
i = r1 + 1
while i <= r2 - 1 and len(ans) < m * n:
ans.append(matrix[i][c2])
i += 1
j = c2
while j >= c1 and len(ans) < m * n:
ans.append(matrix[r2][j])
j -= 1
i = r2 - 1
while i >= r1 + 1 and len(ans) < m * n:
ans.append(matrix[i][c1])
i -= 1
r1 += 1
c1 += 1
r2 -= 1
c2 -= 1

return ans
``````