华为OD机试 - 螺旋数字矩阵（Java & JS & Python & C & C++）

题目描述：

给定一个正整数 n，生成一个包含 1 到 n^2 所有整数的矩阵，但是矩阵是由外向内螺旋状地填充的。

示例：

输入 n = 3

输出

[ 1, 2, 3 ],

[ 8, 9, 4 ],

[ 7, 6, 5 ]

]

解法1：模拟螺旋填充过程（Java）

class Solution {
    public int[][] generateMatrix(int n) {
        int[][] matrix = new int[n][n];
        int start = 0;
        int count = 1;
        int loop = n / 2;
        int mid = n / 2;
        int offset = 1;
        int i, j;
 
        while (loop-- > 0) {
            i = start;
            j = start;
            // left to right
            for (j = start; j < n - offset; j++) {
                matrix[start][j] = count++;
            }
            // top to bottom
            for (i = start; i < n - offset; i++) {
                matrix[i][j] = count++;
            }
            // right to left
            for (; j > start; j--) {
                matrix[i][j] = count++;
            }
            // bottom to top
            for (; i > start; i--) {
                matrix[i][j] = count++;
            }
            start++;
            offset += 1;
        }
 
        if (n % 2 == 1) {
            matrix[mid][mid] = count;
        }
 
        return matrix;
    }
}

解法2：优化空间复杂度（Python）

class Solution:
    def generateMatrix(self, n: int) -> List[List[int]]:
        matrix = [[0] * n for _ in range(n)]
        start = 0
        count = 1
        loop = n // 2
        mid = n // 2
        offset = 1
 
        while loop:
            i = j = start
            # left to right
            for j in range(start, n - offset):
                matrix[start][j] = count
                count += 1
            # top to bottom
            for i in range(start, n - offset):
                matrix[i][j] = count
                count += 1
            # right to left
            for j in range(n - offset, start, -1):
                matrix[i][j] = count
                count += 1
            # bottom to top
            for i in range(n - offset, start, -1):
                matrix[i][j] = count
                count += 1
            start += 1
            offset += 1
            loop -= 1
 
        if n % 2:
            matrix[mid][mid] = count
 
        return matrix

解法3：C++版本

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        vector<vector<in