Leetcode36. 有效的独数

一、题目描述：

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。

1. 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

2. 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 ‘.’

二、解决思路和代码

1. 解决思路

• 分析：这道题统计一下数字 board[i][j] 在第 i 行，第 j 列，以及所在 3x3 宫格中出现的次数，如果所有的数字都只出现一次，那就是一个有效的独数。这里需要考虑一下 3x3 宫格，怎样通过行和列计算出对应的 3x3 宫格？对于 3x3 宫格的第 k 个宫格，$k=i//(row//3)\ast (col//3)+j//3$

2. 代码

from typing import *
class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:        
        x33 = {0:[],1:[],2:[],3:[],4:[],5:[],6:[],7:[],8:[]}
        col = {0:[],1:[],2:[],3:[],4:[],5:[],6:[],7:[],8:[]}
        for i in range(len(board)):
            row = []
            for j in range(len(board[0])):
                if board[i][j]=='.': continue

                if board[i][j] in row: return False
                row.append(board[i][j])
                
                if board[i][j] in col[j]: return False
                col[j].append(board[i][j])
                
                if board[i][j] in x33[i//3*3+j//3]: return False
                x33[i//3*3+j//3].append(board[i][j])
        return True