本文实例讲述了Python判断有效的数独算法。分享给大家供大家参考，具体如下：

一、题目

判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。

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。

数独部分空格内已填入了数字，空白格用 ‘.' 表示。

例1：

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

例2：

输入:

[

["8","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: false

解释: 除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。

但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。

二、解法

先创建三个空数组 row、col、cell，以 cell 为例，里面的每个空字典都代表一个 3×3单元格，然后我们需要把数据一个个填进去

遍历整个二维数组，然后边遍历边把数组分别存入到 行 row , 列 col , 3×3单元格 cell 内的字典，存为key ，而不是 value 。

然后我们就可以判断,行、列、3×3单元格 对应的字典内是否已经存在board[x][y]这个键名，如果存在，那么说明重复了，返回 False

注意，字典中的值这里都为1，但是没有任何意义，你可以随意更改

把数组存入 3×3的单元格是一个难点，num = 3*(x//3)+y//3，这个式子是关键，可以找个数独，然后代入进去好好理解下

当然你也可以不用这个式子，用if/else语句来判断也行，那样比较好理解，但是不如这个式子简洁

类似于: if y<3 : ... elif 3<=y<6 : ... elif 6<=y : ...,

代码如下：

#row，col，cell分别代表行，列，3x3单元格

row, col, cell =

[{}, {}, {}, {}, {}, {}, {}, {}, {}],

[{}, {}, {}, {}, {}, {}, {}, {}, {}],

[{}, {}, {}, {}, {}, {}, {}, {}, {}]

for x in range(9):

for y in range(9):

#取得单元格

num = 3*(x//3)+y//3

temp = board[x][y]

#不需要存入 '.'

if temp != '.':

if (temp not in row[x]

and temp not in col[y]

and temp not in cell[num]):

row[x][temp] = '1'

col[y][temp] = '1'

cell[num][temp] = '1'

else:

return False

return True

时间 64ms，击败了 99.3%

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希望本文所述对大家Python程序设计有所帮助。