Leetcode Spiral Matrix II problem solution

In this Leetcode Spiral Matrix II problem solution we have given a positive integer n, generate an n x n matrix filled with elements from 1 to n2 in spiral order.

Problem solution in Python.

class Solution:
    def generateMatrix(self, n: int) -> List[List[int]]:
        matrix = []
        total = n*n
        for i in range(1,total+1):
            matrix.append(i)
        result = [[None]*n for _ in range(n)]
        
        for d in range(n-1):
            for i in range(d, n-d-1):
                result[d][i]=matrix.pop(0)
            for i in range(d, n-d-1):
                result[i][n-d-1]=matrix.pop(0)
            for i in range(n-d-1,d,-1):
                result[n-d-1][i]=matrix.pop(0)
            for i in range(n-d-1,d,-1):
                result[i][d]=matrix.pop(0)
        if n%2:
            x= (n-1)//2
            result[x][x]=matrix.pop()
        return result

Problem solution in Java.

public int[][] generateMatrix(int n) {
        int[][] directions = new int[][]{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        int dirIndex = 0;
        int x = 0, y = 0;
        int[][] res = new int[n][n];
        for (int i = 1; i <= n * n; i++) {
            res[x][y] = i;
            //System.out.println(Arrays.deepToString(res));
            x = x + directions[dirIndex][0];
            y = y + directions[dirIndex][1];
            if (!isValid(x, y, res, n)) {
                x = x - directions[dirIndex][0];
                y = y - directions[dirIndex][1];
                dirIndex = (dirIndex + 1) % 4;
                x = x + directions[dirIndex][0];
                y = y + directions[dirIndex][1];
            }
        }
        return res;
    }

    private boolean isValid(int x, int y, int[][] res, int n) {
        return (x >= 0 && x < n && y >= 0 && y < n && res[x][y] == 0);
    }

Problem solution in C++.

vector<vector<int> > generateMatrix(int n) {
    vector<vector<int> >matrix(n,vector<int>(n,0));
    int r = n-1;
    int c = n-1;
    int cur = 1;
    for(int i = 0;i<(n+1)/2;++i)
    {
        for(int j = i;j<=c-i;++j)  matrix[i][j] = cur++;
        for(int j = i+1;j<=r-i ;++j)  matrix[j][c-i] = cur++;
        for(int j = c-i-1;j>=i && r-i>i;--j)  matrix[r-i][j] = cur++;
        for(int j = r-i-1;j>i && c-i>i ;--j)  matrix[j][i] = cur++;
    }
    return matrix;
}

Problem solution in C.

int** generateMatrix(int n, int* returnSize, int** returnColumnSizes){
    int direction[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    *returnSize = n;
    int** matrix = (int**) malloc(n * sizeof(int*));
    *returnColumnSizes = (int*) malloc(n * sizeof(int));
    int counter = 1;
    for (int i = 0; i < n; i++) {
        matrix[i] = (int*) malloc(n * sizeof(int));
        (*returnColumnSizes)[i] = n;
    }
    for (int round = 0; round < (n + 1) / 2; round++){
        int x = round, y = round;
        matrix[x][y] = counter;     /* This is for the case n is odd. */
        for (int i = 0; i < 4; i++) {
            for (int j = round; j < n - 1 - round; j++) {
                matrix[x][y] = counter++;
                x += direction[i][0];
                y += direction[i][1];
            }
        }
    }
    return matrix;
}