leetcode 36 有效的數獨

利用二重迴圈的下標來表示行、列和小正方形，減少不必要的資料儲存，每一次內迴圈都代表著對一行或是一列或是乙個小正方形的遍歷。外迴圈則代表對下一行，下一列和下乙個小正方形的遍歷。

數字1-9在每一行只能出現一次。

數字1-9在每一列只能出現一次。

數字1-9在每乙個以粗實線分隔的3x3宮內只能出現一次。

上圖是乙個部分填充的有效的數獨。

數獨部分空格內已填入了數字，空白格用'.'表示。

輸入:

[ ["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"]
]輸出: true

class solution:

def isvalidsudoku(self, board):
""":type board: list[list[str]]
:rtype: bool
"""row = [{},{},{},{},{},{},{},{},{}] # 值：1
col = [{},{},{},{},{},{},{},{},{}]
cell = [{},{},{},{},{},{},{},{},{}]
n = len(board)
for i in range(n):
for j in range(n):
block = 3*(i//3)+j//3 # 找單元
num = board[i][j]
if num != '.':
if num not in row[i] and num not in col[j] and num not in cell[block]:
row[i][num] = 1
col[j][num] = 1
cell[block][num] = 1
else:
return false
# print(row)
# [, , 
# , , 
# , , 
# , , 
# ]return true

leetcode 36 vaild soduku(有效的數獨） python3 最簡**(單次迴圈)

leetcode 36. 有效的數獨 (python)

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