今天是紀錄LeetCode解題的第三十六天

第三十六題題目：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.

判斷9x9數獨是否有效，只有已填充的格子需要根據以下規則進行驗證：

1. 每行必須包含不重複的數字 1-9
2. 每列必須包含不重複的數字 1-9
3. 3x3九宮格中必須包含不重複的數字 1-9

注意：

• 數獨（部分填充）可能有效，但不一定可解
• 只有已填入的格子才需要根據上述規則進行驗證

解題思路

遍歷行，將數字儲存至串列中，判斷串列和集合(把串列轉成集合)個數是否相同，如果不同，代表該行有重複元素，直接回傳False，表示該數獨無效，列、九宮格同理

程式碼

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        for i in range(9):
            row_nums = []
            col_nums = []
            for j in range(9):
                if board[i][j] != ".":
                    row_nums.append(board[i][j])
                if board[j][i] != ".":
                    col_nums.append(board[j][i])
            if len(row_nums) != len(set(row_nums)):
                return False
            if len(col_nums) != len(set(col_nums)):
                return False
        for block_row in range(3):
            for block_col in range(3):
                nine_nums = []
                for i in range(3):
                    for j in range(3):
                        row = block_row * 3 + i
                        col = block_col * 3 + j
                        if board[row][col] != ".":
                            nine_nums.append(board[row][col])
                if len(nine_nums) != len(set(nine_nums)):
                    return False
        return True

這題比較簡單，只是判斷數獨有無效，只要檢查是否有重複項即可