前言

大家好，我是毛毛。ヾ(´∀ ˋ)ﾉ
那就開始今天的解題吧~

Question

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

1. Each row must contain the digits 1-9 without repetition.
2. Each column must contain the digits 1-9 without repetition.
3. 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 1:

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

Example 2:

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 '.'.

Think

簡單來說就是判斷是不是合乎規則的數獨~

分別判斷行、列跟九塊九宮格是否都有符合只出現一次的1~9

Code

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

Submission

今天就到這邊啦~
大家明天見