🧠 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：
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
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 Falsebox_r, box_c = 3 * (r // 3) + i // 3, 3 * (c // 3) + i % 3if board[box_r][box_c] == ch:return Falsereturn Truedef 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] = chif dfs():return Trueboard[i][j] = "."return Falsereturn Truedfs()
📌 四、输出示例（运行后 board 被原地修改）
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！
