【LeetCode】36. 有效的数独

1 问题

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。

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"]]
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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"]]
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8
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。

2 答案

这题直接不会
暴力法，自己写的，速度比较快

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        
        for i in range(len(board)):
            hashmap = {}
            for j in range(len(board[0])):
                if board[i][j] != '.':
                    hashmap[board[i][j]] = hashmap.get(board[i][j], 0) + 1
                    if hashmap[board[i][j]] > 1:
                        return False
        
        for j in range(len(board[0])):
            hashmap = {}
            for i in range(len(board)):
                if board[i][j] != '.':
                    hashmap[board[i][j]] = hashmap.get(board[i][j], 0) + 1
                    if hashmap[board[i][j]] > 1:
                        return False
        
        
        for i in range(3):
            for j in range(3):
                hashmap = {} 
                for m in range(len(board)):
                    for n in range(len(board[0])):
                        if board[m][n] != '.' and m // 3 == i and n // 3 == j:
                            hashmap[board[m][n]] = hashmap.get(board[m][n], 0) + 1
                            if hashmap[board[m][n]] > 1:
                                return False  
        return True
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官方解，利用哈希表，哈希表也可以是列表，矩阵的表示方法也很妙，非常值得学习

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:

        row = [[0] * 9 for _ in range(9)]
        col = [[0] * 9 for _ in range(9)]
        block = [[0] * 9 for _ in range(9)]

        for i in range(9):
            for j in range(9):
                if board[i][j] != '.':
                    num = int(board[i][j]) - 1
                    b = (i // 3) * 3 + j // 3   # 3*3 小矩阵的第b个矩阵，这个很妙
                    if row[i][num] or col[j][num] or block[b][num]:  # 看这几个位置是否有值，如果有一个有，则返回 False
                        return False
                    row[i][num] = col[j][num] = block[b][num] = 1
        return True
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3 知识点

哈希表（Hash Table）：也叫做散列表。是根据关键码值（Key Value）直接进行访问的数据结构。

哈希表通过「键 key 」和「映射函数 Hash(key) 」计算出对应的「值value」，把关键码值映射到表中一个位置来访问记录，以加快查找的速度。这个映射函数叫做「哈希函数（散列函数）」，存放记录的数组叫做「哈希表（散列表）」。

https://blog.csdn.net/zy_dreamer/article/details/131036258