一、题目描述:
请你判断一个 9 x 9 的数独是否有效。只需要 根据以下规则 ,验证已经填入的数字是否有效即可。
- 数字 1-9 在每一行只能出现一次。
- 数字 1-9 在每一列只能出现一次。
- 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。(请参考示例图)
注意:
- 一个有效的数独(部分已被填充)不一定是可解的。
- 只需要根据以上规则,验证已经填入的数字是否有效即可。
- 空白格用 ‘.’ 表示。
-
示例 1:
- 输入:
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"]]
- 输出:true
- 输入:
-
示例 2:
- 输入:
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"]]
- 输出:false
- 解释:除了第一行的第一个数字从 5 改为 8 以外,空格内其他数字均与 示例1 相同。 但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。
- 输入:
- 提示:
- board.length == 9
- board[i].length == 9
- board[i][j] 是一位数字(1-9)或者 ‘.’
二、解决思路和代码
1. 解决思路
- 分析:这道题统计一下数字 board[i][j] 在第 i 行,第 j 列,以及所在 3x3 宫格中出现的次数,如果所有的数字都只出现一次,那就是一个有效的独数。这里需要考虑一下 3x3 宫格,怎样通过行和列计算出对应的 3x3 宫格?对于 3x3 宫格的第 k 个宫格, k = i / / ( r o w / / 3 ) ∗ ( c o l / / 3 ) + j / / 3 k=i//(row//3)*(col//3)+j//3 k=i//(row//3)∗(col//3)+j//3
2. 代码
from typing import *
class Solution:
def isValidSudoku(self, board: List[List[str]]) -> bool:
x33 = {
0:[],1:[],2:[],3:[],4:[],5:[],6:[],7:[],8:[]}
col = {
0:[],1:[],2:[],3:[],4:[],5:[],6:[],7:[],8:[]}
for i in range(len(board)):
row = []
for j in range(len(board[0])):
if board[i][j]=='.': continue
if board[i][j] in row: return False
row.append(board[i][j])
if board[i][j] in col[j]: return False
col[j].append(board[i][j])
if board[i][j] in x33[i//3*3+j//3]: return False
x33[i//3*3+j//3].append(board[i][j])
return True