有效的数独Python解法

请你判断一个 9 x 9 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。

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）

注意：

一个有效的数独（部分已被填充）不一定是可解的。
只需要根据以上规则，验证已经填入的数字是否有效即可。
空白格用 '.' 表示。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/valid-sudoku

例：

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：true

解析：

按题目要求分三个部分判断，分别为行，列和块即可。

class Solution(object):def isValidSudoku(self, board):""":type board: List[List[str]]:rtype: bool"""for i in range(0, 9):  # 外圈遍历row = []  # 行判断数组col = []  # 列判断数组block = []  # 块判断数组for j in range(0, 9):  # 内圈遍历if board[i][j] == '.' or board[i][j] not in row:  # 行判断row.append(board[i][j])else:return Falseif board[j][i] == '.' or board[j][i] not in col:  # 列判断col.append(board[j][i])else:return Falseblock_i = 3 * (i / 3) + j / 3  # 总行 = 大行加小行block_j = 3 * (i % 3) + j % 3  # 总列 = 大列加小列if board[block_i][block_j] == '.' or board[block_i][block_j] not in block:  # 块判断block.append(board[block_i][block_j])else:return Falsereturn True

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