分类：HashTable
考察知识点：HashTable 数组遍历
最优解时间复杂度：O(n^2)
36. Valid Sudoku
Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:
- Each row must contain the digits
1-9without repetition. - Each column must contain the digits
1-9without repetition. - Each of the 9
3x3sub-boxes of the grid must contain the digits1-9without repetition.
A partially filled sudoku which is valid.
The Sudoku board could be partially filled, where empty cells are filled with the character '.'.
Example 1:
Input:
[
["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"]
]
Output: true
Example 2:
Input:
[
["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"]
]
Output: false
Explanation: Same as Example 1, except with the 5 in the top left corner being
modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.
Note:
- A Sudoku board (partially filled) could be valid but is not necessarily solvable.
- Only the filled cells need to be validated according to the mentioned rules.
- The given board contain only digits
1-9and the character'.'. - The given board size is always
9x9.
代码：
HashMap解法O(n^2):
class Solution:
def isValidSudoku(self, board):
"""
:type board: List[List[str]]
:rtype: bool
"""
for i in range(9):
#初始化三个dict
row_dict={}
col_dict={}
cube_dict={}
for j in range(9):
#判断横行
if board[i][j]!=".":
if board[i][j] in row_dict:
return False
else:
row_dict[board[i][j]]=1
#判断纵行
if board[j][i]!=".":
if board[j][i] in col_dict:
return False
else:
col_dict[board[j][i]]=1
#判断Cube
rowIndex=(i//3)*3
colIndex=(i%3)*3
if board[rowIndex+j//3][colIndex+j%3]!=".":
if board[rowIndex+j//3][colIndex+j%3] in cube_dict:
return False
else:
cube_dict[board[rowIndex+j//3][colIndex+j%3]]=1
return True
普通解法O(n^3):
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]!=".":
if (self.isValid(board,i,j))==False:
return False
return True
def isValid(self,board,row,col):
#检验横行和纵行
for i in range(9):
if row!=i:
if board[i][col]==board[row][col]:
return False
if col!=i:
if board[row][i]==board[row][col]:
return False
#检验cube
for i in range(row//3*3,(row//3+1)*3):
for j in range(col//3*3,(col//3+1)*3):
if i!=row and j!=col:
if board[i][j]==board[row][col]:
return False
return True
讨论：
1.这道题目非常重要，在公司的面试中出现的概率特别的高
2.有两种写法，一种是用HashMap，一种就是普通判断
3.对于这个i//3和i%3的方法可以熟练一下，比较取巧，但是还是可以用的！
一般解速度
最优解速度
