Day 10 - Valid Sudoku

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tags: `2021_iThome铁人赛` `Leetcode`

36. Valid Sudoku

Question

Determine if a 9 x 9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

Each row must contain the digits 1-9 without repetition.
Each column must contain the digits 1-9 without repetition.
Each of the nine 3 x 3 sub-boxes of the grid must contain the digits 1-9 without repetition.

Note:

A Sudoku board (partially filled) could be valid but is not necessarily solvable.
Only the filled cells need to be validated according to the mentioned rules.

Example

Example1

Input: 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"]]
Output: true

Example2

Input: 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"]]
Output: false
Explanation: Same as Example 1, except with the 5 in the top left corner being modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.

Constraints

board.length == 9
board[i].length == 9
board[i][j] is a digit 1-9 or '.'.

解题

题目

首先先简单的翻译一下题目
~~给一个数独题目要你解~~，给一个数独题目要判断这个数独题目是不是合乎数独的规则，也就是不管是行、列和9块3 x 3的区块，出现的数字1-9皆不能重复。

Think

作法大致上是这样

本来一开始是打算用for loop做，但是忽然想到有enumerate可以使用，enumerate有点像for rowVal in board:的方式，只是每个值会在额外配上一个sequence，这个sequence的起始可以透过start参数来决定。
因为有行、列和区块要判断，所以就分三个部份分别处理。
下面避免行列混淆，用Row跟Column来说明。
Row part & Column part
- 由row和col这两个list存着已经出现过的值，假如新读到的值已经有存在里头了，就代表重复了，回传False；没有的话，就存入。
- 存入的值为(seq, value)搭配
subBox part
- 由subBox这个list存着已经出现过的值，假如新读到的值已经有存在里头了，就代表重复了，回传False；没有的话，就存入。
- 存入的值是取seq/3的商，这边我用转int把小数部分舍去了。
- 这样可以得到9个区块分别的代码：
  - 0,0、0,1、0,2
  - 1,0、1,1、1,2
  - 2,0、2,1、2,2
- 在配上每个点的value就可以透过subBox的list判断有没有重复数字了。
C的程序出了点问题QQ，再补上。

Code

Python

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        row = []
        col = []
        subBox = []
        
        for seq1, rowVal in enumerate(board, start=0):
            for seq2, colVal in enumerate(rowVal, start=0):

                if colVal is not ".":
                    # Row part
                    if (seq1, colVal) not in row:
                        row.append((seq1, colVal))
                    else:
                        return False

                    # Column part
                    if (seq2, colVal) not in col:
                        col.append((seq2, colVal))
                    else:
                        return False
                
                    # subBox part
                    if (int(seq1/3), int(seq2/3), colVal) not in subBox:
                        subBox.append((int(seq1/3), int(seq2/3), colVal))
                    else:
                        return False
        return True