LeeCode题库第三十七题

        37.解数独 

项目场景：

编写一个程序，通过填充空格来解决数独问题。

数独的解法需 遵循如下规则：

1. 数字 1-9 在每一行只能出现一次。
2. 数字 1-9 在每一列只能出现一次。
3. 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。（请参考示例图）

数独部分空格内已填入了数字，空白格用 '.' 表示。

示例 1：

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"]]
输出：[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
解释：输入的数独如上图所示，唯一有效的解决方案如下所示：

提示：

• board.length == 9
• board[i].length == 9
• board[i][j] 是一位数字或者 '.'
• 题目数据 保证 输入数独仅有一个解

---

问题描述

        这段代码实现了一个 数独求解器，使用深度优先搜索（DFS）和回溯算法来填充数独棋盘中的空白格。代码的核心逻辑是通过递归尝试在每个空白格中填入数字 1 到 9，并利用三个辅助数据结构（line、column 和 block）来高效检查数字是否可用。如果填入的数字导致后续无解，则回溯并尝试其他数字；如果所有空白格都被正确填充，则设置 valid = True 并提前返回，避免不必要的计算。最终，代码会修改输入的 board，将其填充为一个有效的数独解。

class Solution:def solveSudoku(self, board: List[List[str]]) -> None:def dfs(pos:int):nonlocal validif pos==len(spaces):valid=Truereturn i,j=spaces[pos]for digit in range(9):if line[i][digit]==column[j][digit]==block[i//3][j//3][digit]==False:line[i][digit]=column[j][digit]=block[i//3][j//3][digit]=Trueboard[i][j]=str(digit+1)dfs(pos+1)line[i][digit]=column[j][digit]=block[i//3][j//3][digit]=Falseif valid:returnline=[[False]*9 for _ in range(9)]column=[[False]*9 for _ in range(9)]block=[[[False]*9 for _a in range(9)]for _b in range(9)]valid=Falsespaces=[]for i in range(9):for j in range(9):if board[i][j]=='.':spaces.append((i,j))else:digit=int(board[i][j])-1line[i][digit]=column[j][digit]=block[i//3][j//3][digit]=Truedfs(0)

        本题提交情况。

 

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