Likou NO.36 Effective Sudoku (Hash Table)

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Solution 1: Three hash tables store the number of occurrences of the corresponding position number

Hash table of row, column, box
row_haxi = [[0] * 10 for _ in range(9)] # Take num = int(board[i][j]) Let row_haxi[i][num]+=1
column_haxi = [[0] * 10 for _ in range(9)] # Take num = int(board[i][j]) so that column_haxi[j][num]+=1
box_haxi = [[0] * 10 for _ in range(9)] # Take num = int(board[i][j]) m(number of boxes): m = j//3 + i//3*3;n(number of boxes) : n = num let box_haxi[m][n]+=1

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        # 行，列，box 的哈希表
        row_haxi = [[0] * 10 for _ in range(9)] # 取num = int(board[i][j]) 令 row_haxi[i][num]+=1
        column_haxi = [[0] * 10 for _ in range(9)] # 取num = int(board[i][j]) 令 column_haxi[j][num]+=1
        box_haxi = [[0] * 10 for _ in range(9)] # 取num = int(board[i][j]) m(第几个box)：m = j//3 + i//3*3;n(第几个数)：n = num 令 box_haxi[m][n]+=1
        for i in range(9):
            for j in range(9):
                if board[i][j]!='.':  # 非空值时进行判断
                    num = int(board[i][j])
                    row_haxi[i][num]+=1   
                    column_haxi[j][num]+=1
                    box_haxi[j//3 + i//3*3][num]+=1
                    # 判断是否合规
                    if row_haxi[i][num]>1:return False
                    if column_haxi[j][num]>1:return False
                    if box_haxi[j//3 + i//3*3][num]>1:return False
        return True