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%
希望本文所述对大家Python程序设计有所帮助。
Python判断有效的数独算法示例
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