用 Python 编写程序来验证数独网格是否可解
pythonserver side programmingprogramming更新于 2026/2/4 3:56:17
假设我们有一个 9×9 数独网格。我们现在必须检查它是否有效。只需根据以下规则验证已填充的单元格 −
每行必须包含从 1−9 开始的数字,且不能重复。
每列必须包含从 1−9 开始的数字,且不能重复。
网格的 9 个(3−3)子框中的每一个都必须包含从 1−9 开始的数字,且不能重复。
假设数独网格就像 −

这是有效的。
为了解决这个问题,我们将遵循以下步骤 −
对于 i 在 0 到 8 范围内
创建一些名为 row、col 和 block 的空字典,row_cube := 3 * (i / 3),col_cube := 3 * (i mod 3)
对于 j 在 0 到 8 范围内
如果 board[i, j] 不为空且 board[i, j] 在 row 中,则返回 false
row[board[i, j]] := 1
如果 board[j, i] 不为空且 board[j, i] 在 col 中,则返回 false
col[board[j, i]] := 1
rc := row_cube + j/3 和 cc := col_cube + j mod 3
如果 board[rc, cc] 在 block 中且 board[rc, cc] 不为空,则返回 false
block[board[rc, cc]] := 1
返回 true
让我们看看下面的实现以便更好地理解 −
示例
class Solution(object):
def isValidSudoku(self, board):
for i in range(9):
row = {}
column = {}
block = {}
row_cube = 3 * (i//3)
column_cube = 3 * (i%3)
for j in range(9):
if board[i][j]!='.' and board[i][j] in row:
return False
row[board[i][j]] = 1
if board[j][i]!='.' and board[j][i] in column:
return False
column[board[j][i]] = 1
rc= row_cube+j//3
cc = column_cube + j%3
if board[rc][cc] in block and board[rc][cc]!='.':
return False
block[board[rc][cc]]=1
return True
ob1 = Solution()
print(ob1.isValidSudoku([
["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"]]))
输入
[ ["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
True
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有用资源
python 参考教程 - 该教程包含有关 python 的更多信息:https://www.cainiaomax.com/python/

