# Spiral Matrix II Leetcode Solution In this post, you will know how to solve the Spiral Matrix II 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 a positive integer `n`, generate an `n x n` `matrix` filled with elements from `1` to `n2` in spiral order.

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

Example 2:

```Input: n = 1
Output: []
```

Constraints:

• `1 <= n <= 20`

### Spiral Matrix II Leetcode Solutions in Python

```class Solution:
def generateMatrix(self, n: int) -> List[List[int]]:
ans = [ * n for _ in range(n)]
count = 1
for min in range(n // 2):
max = n - min - 1
for i in range(min, max):
ans[min][i] = count
count += 1
for i in range(min, max):
ans[i][max] = count
count += 1
for i in range(max, min, -1):
ans[max][i] = count
count += 1
for i in range(max, min, -1):
ans[i][min] = count
count += 1
if n & 1:
ans[n // 2][n // 2] = count
return ans
```

### Spiral Matrix II Leetcode Solutionsin CPP

```class Solution {
public:
vector<vector<int>> generateMatrix(int n) {
vector<vector<int>> ans(n, vector<int>(n));
int count = 1;
for (int min = 0; min < n / 2; ++min) {
const int max = n - min - 1;
for (int i = min; i < max; ++i)
ans[min][i] = count++;
for (int i = min; i < max; ++i)
ans[i][max] = count++;
for (int i = max; i > min; --i)
ans[max][i] = count++;
for (int i = max; i > min; --i)
ans[i][min] = count++;
}
if (n & 1)
ans[n / 2][n / 2] = count;
return ans;
}
};
```

### Spiral Matrix II Leetcode Solutions in Java

```class Solution {
public int[][] generateMatrix(int n) {
int[][] ans = new int[n][n];
int count = 1;
for (int min = 0; min < n / 2; ++min) {
final int max = n - min - 1;
for (int i = min; i < max; ++i)
ans[min][i] = count++;
for (int i = min; i < max; ++i)
ans[i][max] = count++;
for (int i = max; i > min; --i)
ans[max][i] = count++;
for (int i = max; i > min; --i)
ans[i][min] = count++;
}
if (n % 2 == 1)
ans[n / 2][n / 2] = count;
return ans;
}
}
```

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