# Spiral Matrix Leetcode Solution

#### ByBrokenprogrammers

Dec 17, 2022

In this post, you will know how to solve the Spiral Matrix Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.

## Problem

Given an `m x n` `matrix`, return all elements of the `matrix` in spiral order.

Example 1:

```Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]
```

Example 2:

```Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]
```

Constraints:

• `m == matrix.length`
• `n == matrix[i].length`
• `1 <= m, n <= 10`
• `-100 <= matrix[i][j] <= 100`

### Spiral Matrix Leetcode Solutions in 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
```

### Spiral Matrix Leetcode Solutions in CPP

```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;
}
};
```

### Spiral Matrix Leetcode Solutions in 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)
for (int i = r1 + 1; i <= r2 - 1 && ans.size() < m * n; ++i)
for (int j = c2; j >= c1 && ans.size() < m * n; --j)
for (int i = r2 - 1; i >= r1 + 1 && ans.size() < m * n; --i)
++r1;
++c1;
--r2;
--c2;
}
return ans;
}
}
```

Note: This problem Spiral Matrix is generated by Leetcode but the solution is provided by  BrokenProgrammers. This tutorial is only for Educational and Learning purposes.