有效的数独
判断一个 9x9 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。
上图是一个部分填充的有效的数独。
数独部分空格内已填入了数字,空白格用 '.'
表示。
示例 1:
输入:
[
["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"]
]
输出: true
示例 2:
输入:
[
["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。
说明:
- 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。
- 给定数独序列只包含数字
1-9
和字符'.'
。 - 给定数独永远是
9x9
形式的
这道题第一眼看挺懵的,感觉只能行、列、3*3块分别检测,后来想到3*3块就没办法了
从网上寻找答案,将索引到位置利用整数除法框定范围判定是否存在相同的数字
v1.0时间复杂度O(n^3)
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class Solution:
def isValidSudoku(self, board):
"""
:type board: List[List[str]]
:rtype: bool
"""
for i in range(9):
for j in range(9):
if board[i][j]!= '.':
#判断行重复性
for k in range(j+1,9):
if board[i][j] == board[i][k]:
return False
for h in range(i+1,9):
if board[i][j] == board[h][j]:
return False
a = board[i][j]
board[i][j] = 0
for p in range(i//3*3, i//3*3+3): #//返回不大于结果的最大整数
for q in range(j//3*3, j//3*3+3):
if a== board[p][q]:
return False
return True
时间少用了很多,减少了数据读取
def isValidSudoku(board):
dic_row = [{},{},{},{},{},{},{},{},{}]
dic_column =[{},{},{},{},{},{},{},{},{}]
dic_box = [{},{},{},{},{},{},{},{},{}]
for i in range(9):
for j in range(9):
num = board[i][j]
if num == '.':
continue
if num not in dic_row[i] and num not in dic_column[j] and num not in dic_box[3*(i//3)+(j//3)]:
#判定key是否存在
dic_row[i][num] =1
dic_column[j][num] = 1
dic_box[3*(i//3)+(j//3)][num] = 1
else:
return False
return True
利用字典的方法,每一行即为一个字典,将第i行或者第j列的key= num 时设为1
判定若num存在相同的key值,即为存在重复
v3.0 v2.0另一种实现方法
def isValidSudoku(board):
Cell = [[] for i in range(9)] # 没有必要用dict,我们只某个数字关心有没有出现过
Col = [[] for i in range(9)]
Row = [[] for i in range(9)]
for i,row in enumerate(board): # 将一个可遍历的数据对象(如列表、元组或字符串)组合为一个索引序列,同时列出数据和数据下标,一般用在 for 循环当中
for j,num in enumerate(row):
if num != '.':
k = (i//3)*3 + j//3
if num in Row[i] + Col[j] + Cell[k]: # list的骚操作,将三个list顺序的拼接
return False
Row[i].append(num)
Col[j].append(num)
Cell[k].append(num)
return True
设定行列与块cell分别为9个
for i,row in enumerate(board): # 将一个可遍历的数据对象(如列表、元组或字符串)组合为一个索引序列,同时列出数据和数据下标,一般用在 for 循环当中
for j,num in enumerate(row):
同时输出输入对象的行列和结果值进行处理
判定num是否存在于与Row[i] + Col[j] + Cell[k]中
不存在之后执行调加
存在直接返回False