Python判断有效数独算法的示例-创新互联
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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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