Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.
The Sudoku board could be partially filled, where empty cells are filled with the character ‘.‘
.
A partially filled sudoku which is valid.
Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.
验证当前九宫格盘面是否有错误
class Solution:
def valid3by3(self, board, row, col):
validPos = [
[-1, -1], [-1, 0], [-1, 1],
[0, -1], [0, 0], [0, 1],
[1, -1], [1, 0], [1, 1]
]
s = set()
for pos in validPos:
curVal = board[row + pos[0]][col + pos[1]]
if curVal is ".":
continue
if curVal in s:
return False
s.add(curVal)
return True
def validRow(self, board, row):
s = set()
for curVal in board[row]:
if curVal is ".":
continue
if curVal in s:
return False
s.add(curVal)
return True
def validCol(self, board, col):
s = set()
for row in board:
curVal = row[col]
if curVal is ".":
continue
if curVal in s:
return False
s.add(curVal)
return True
def isValidSudoku(self, board):
"""
:type board: List[List[str]]
:rtype: bool
"""
pos3by3 = [
[1, 1], [1, 4], [1, 7],
[4, 1], [4, 4], [4, 7],
[7, 1], [7, 4], [7, 7]
]
for pos in pos3by3:
if not self.valid3by3(board, pos[0], pos[1]):
return False
for row in range(0, 9):
if not self.validRow(board, row):
return False
for col in range(0, 9):
if not self.validCol(board, col):
return False
return True
s = Solution()
board = [
[".", "8", "7", "6", "5", "4", "3", "2", "1"],
["2", ".", ".", ".", ".", ".", ".", ".", "."],
["3", ".", ".", ".", ".", ".", ".", ".", "."],
["4", ".", ".", ".", ".", ".", ".", ".", "."],
["5", ".", ".", ".", ".", ".", ".", ".", "."],
["6", ".", ".", ".", ".", ".", ".", ".", "."],
["7", ".", ".", ".", ".", ".", ".", ".", "."],
["8", ".", ".", ".", ".", ".", ".", ".", "."],
["9", ".", ".", ".", ".", ".", ".", ".", "."]
]
res = s.isValidSudoku(board)
print(res)
原文地址:https://www.cnblogs.com/xiejunzhao/p/8319163.html
时间: 2024-11-08 15:17:52