给你一个正整数 n ，生成一个包含 1 到 n2 所有元素，且元素按顺时针顺序螺旋排列的 n x n 正方形矩阵 matrix 。

示例 1：

输入：n = 3
输出：[[1,2,3],[8,9,4],[7,6,5]]

示例 2：

输入：n = 1
输出：[[1]]

提示：

• 1 <= n <= 20
• 使用和二分法一样的思想，确定循环不变量，确定是左闭右开，还是左闭右闭合，本次使用的是左闭右开

C++实现:

#include "array_algorithm.h"

vector<vector<int>> generateMatrix(int n) {
    int loop_num = n / 2; // 螺旋循环的圈数
    int start_x = 0; // 矩阵行的起始位置 
    int start_y = 0; // 矩阵列的起始位置
    int offset = 1; // 选用左开右闭原则，偏移
    int count = 1; // 赋值参数
    vector<vector<int>> nums(n, vector<int>(n, 0)); // 构建二维数组，全赋值为0
    int i;
    int j;
    while (loop_num--)
    {
        i = start_x;
        j = start_y;

        for (j = start_y; j < n - offset; j++) {
            nums[start_y][j] = count++;
        }
        for (i = start_x; i < n - offset; i++) {
            nums[i][j] = count++;
        }
        for (; j > start_y; j--) {
            nums[i][j] = count++;
        }
        for (; i > start_x; i--) {
            nums[i][j] = count++;
        }

        start_x++;
        start_y++;

        offset += 1;
    }
    if (n % 2 == 1) {
        nums[n / 2][n / 2] = count;
    }
    return nums;
}

python实现：

def generateMatrix(n: int):
    loop_nums = int(n // 2)
    start_x = 0
    start_y = 0
    offset = 1
    count = 1
    # nums = np.random.randint(low=0, high=1, size=(n , n))
    nums = [[0] * n for _ in range(n)]
    for _ in range(loop_nums):
        for i in range(start_y, n - offset):
            nums[start_x][i] = count
            count += 1
        for i in range(start_x, n - offset):
            nums[i][n - offset] = count
            count += 1
        for i in range(n - offset, start_y, -1):
            nums[n - offset][i] = count
            count += 1
        for i in range(n - offset, start_x, -1):
            nums[i][start_y] = count
            count += 1
        start_x += 1
        start_y += 1
        offset += 1
    if n % 2 == 1:
        nums[n//2][n//2] = count
    return nums