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:
- 数字
1-9
在每一行只能出现一次。 - 数字
1-9
在每一列只能出现一次。 - 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 valid
if pos==len(spaces):
valid=True
return
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]=True
board[i][j]=str(digit+1)
dfs(pos+1)
line[i][digit]=column[j][digit]=block[i//3][j//3][digit]=False
if valid:
return
line=[[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=False
spaces=[]
for i in range(9):
for j in range(9):
if board[i][j]=='.':
spaces.append((i,j))
else:
digit=int(board[i][j])-1
line[i][digit]=column[j][digit]=block[i//3][j//3][digit]=True
dfs(0)
本题提交情况。
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