Python描述 LeetCode 36. 有效的数独

Python描述 LeetCode 36. 有效的数独

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题目

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。

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。（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。

示例 1：

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

示例 2：

输入：board = 
[["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。

提示：

• ​​board.length == 9​​
• ​​board[i].length == 9​​
• ​​board[i][j]​​​ 是一位数字（​​1-9​​​）或者​​'.'​​

Python描述

一次遍历，分别使用列表记录，如果有重复的，则返回FALSE

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        line = [[0 for _ in range(10)] for __ in range(9)]
        col = [[0 for _ in range(10)] for __ in range(9)]
        jiu = [[[0 for _ in range(10)] for __ in range(3)] for ___ in range(3)]

        for i in range(9):
            for j in range(9):
                tmp = int(board[i][j]) if board[i][j] != '.' else 0
                if tmp == 0:
                    continue
                
                if line[i][tmp] == 1:
                    return False
                line[i][tmp] += 1

                if col[j][tmp] == 1:
                    return False
                col[j][tmp] += 1

                if jiu[i//3][j//3][tmp] == 1:
                    return False
                jiu[i//3][j//3][tmp] += 1
        return True