Python小练习系列 Vol.5：数独求解（经典回溯 + 剪枝）

滴答滴答嗒嗒滴2025-04-01 16:04

🧠 Python小练习系列 Vol.5：数独求解（经典回溯 + 剪枝）

🧩 数独不仅是益智游戏，更是回溯算法的典范！本期我们将用 DFS + 剪枝的方式一步步求解一个标准 9x9 数独。

🧩 一、题目描述

给定一个部分填充的 9×9 数独棋盘，请编写一个程序将其填完，使每行、每列、每个 3×3 宫内的数字 1~9 均不重复。

示例输入：

python 复制代码

board = [
  ["5","3",".",".","7",".",".",".","."],
  ["6",".",".","1","9","5",".",".","."],
  [".","9","8",".",".",".",".","6","."],
  ["8",".",".",".","6",".",".",".","3"],
  ["4",".",".","8",".","3",".",".","1"],
  ["7",".",".",".","2",".",".",".","6"],
  [".","6",".",".",".",".","2","8","."],
  [".",".",".","4","1","9",".",".","5"],
  [".",".",".",".","8",".",".","7","9"]
]

🧠 二、解题思路

使用回溯 + 剪枝，依次尝试将数字 1~9 填入空格中，若不符合数独规则则回退重试。

核心逻辑：

递归寻找空白格，尝试填入合法数字
若数字在当前行、列或 3x3 宫格中已出现，则剪枝
全部填满时即为解

👨‍💻 三、Python代码实现

python 复制代码

def solve_sudoku(board):
    def is_valid(r, c, ch):
        for i in range(9):
            if board[r][i] == ch or board[i][c] == ch:
                return False
            box_r, box_c = 3 * (r // 3) + i // 3, 3 * (c // 3) + i % 3
            if board[box_r][box_c] == ch:
                return False
        return True

    def dfs():
        for i in range(9):
            for j in range(9):
                if board[i][j] == ".":
                    for ch in map(str, range(1, 10)):
                        if is_valid(i, j, ch):
                            board[i][j] = ch
                            if dfs():
                                return True
                            board[i][j] = "."
                    return False
        return True

    dfs()

📌 四、输出示例（运行后 board 被原地修改）

python 复制代码

solve_sudoku(board)
for row in board:
    print(" ".join(row))

🧩 五、关键点总结

步骤	说明
查找空位	使用双层循环寻找 `.`
合法判断	检查当前行、列、宫格是否冲突
回溯回退	无法填充时，重置该位置为 `.`

✅ 本题是典型的「排列填空 + 剪枝」模型

💡 六、进阶挑战

🧠 输出所有可能解（需去掉第一个 return）
⚡ 加速优化：用位运算预处理可选值
🎨 制作 GUI 数独求解器（Tkinter/PyQt）

❤️ 结语

数独是一道优雅的全排列题目，透过回溯 + 剪枝，掌握"选择-尝试-回退"的算法核心！

📌 下一期预告：单词搜索（网格回溯）

👉 点个赞 👍 + 收藏 🌟，学透回溯，从数独开始！