尝试在Python中创建具有随机数的数独矩阵

import random

def generate_sudoku():
    sudoku = [[0] * 9 for _ in range(9)]  # 创建一个9x9的二维列表，初始值为0
    
    # 填充数独矩阵的对角线区域
    for i in range(0, 9, 3):
        nums = [1, 2, 3, 4, 5, 6, 7, 8, 9]
        random.shuffle(nums)
        for j in range(3):
            sudoku[i+j][i+j] = nums[j]
    
    # 使用回溯法填充数独矩阵的剩余空格
    solve_sudoku(sudoku)
    
    return sudoku

def solve_sudoku(sudoku):
    empty_pos = find_empty_position(sudoku)
    if not empty_pos:
        return True  # 数独已经填满，解决完成
    
    row, col = empty_pos
    random.shuffle(range(1, 10))  # 随机打乱1到9的顺序
    for num in range(1, 10):
        if is_valid_move(sudoku, row, col, num):
            sudoku[row][col] = num
            if solve_sudoku(sudoku):
                return True
            sudoku[row][col] = 0  # 回溯，将当前位置重置为0
    
    return False  # 无解

def is_valid_move(sudoku, row, col, num):
    # 检查行是否合法
    for i in range(9):
        if sudoku[row][i] == num:
            return False
    
    # 检查列是否合法
    for i in range(9):
        if sudoku[i][col] == num:
            return False
    
    # 检查3x3的小方格是否合法
    start_row = (row // 3) * 3
    start_col = (col // 3) * 3
    for i in range(3):
        for j in range(3):
            if sudoku[start_row+i][start_col+j] == num:
                return False
    
    return True

def find_empty_position(sudoku):
    for i in range(9):
        for j in range(9):
            if sudoku[i][j] == 0:
                return (i, j)
    return None

sudoku = generate_sudoku()
for row in sudoku:
    print(row)