Leetcode 36. Valid Sudoko

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.

 

思路： 检查每一行，每一列，每一个3*3的submatrix是不是出现了重复的数字，如果出现了，return False。如果没有renturn True。 L21-L23在Python 3中已经可以省略，如果A是一个list of lists, ''.join(A) 可以直接返回一个string。但是python2中还需要先将A变成一个list。

Tip: 在一个lists of list中如何取出一个submatirx? 假设 row = 1:4, col = 2:5。 可以使用

A = [row[2:5] for row in M[1:4]]

 

 1 class Solution(object):
 2     def isValidSudoku(self, board):
 3         """
 4         :type board: List[List[str]]
 5         :rtype: bool
 6         """
 7         for i in range(9):
 8             if self.isValidString(board[i]) == False:
 9                 return False
10         
11         
12         for i in range(9):
13             string = [board[j][i] for j in range(9)]
14             string = ''.join(string)
15             if self.isValidString(string) == False:
16                 return False
17         
18         for i in range(3):
19             for j in range(3):
20                 string = [row[3*j: 3*j + 3] for row in board[3*i: 3*i+3]]
21                 ans = []
22                 for k in range(len(string)):
23                     ans += string[k]
24                 string = ''.join(ans)
25                 if self.isValidString(string) == False:
26                     return False
27         return True
28         
29     def isValidString(self, numbers):
30         isused = [0] * 10
31         for j in range(9):
32             if numbers[j] == '1' and isused[1] == 0:
33                 isused[1] = 134             elif numbers[j] == '2' and isused[2] == 0:
35                 isused[2] = 136             elif numbers[j] == '3' and isused[3] == 0:
37                 isused[3] = 138             elif numbers[j] == '4' and isused[4] == 0:
39                 isused[4] = 140             elif numbers[j] == '5' and isused[5] == 0:
41                 isused[5] = 142             elif numbers[j] == '6' and isused[6] == 0:
43                 isused[6] = 144             elif numbers[j] == '7' and isused[7] == 0:
45                 isused[7] = 146             elif numbers[j] == '8' and isused[8] == 0:
47                 isused[8] = 148             elif numbers[j] == '9' and isused[9] == 0:
49                 isused[9] = 150             elif numbers[j] == '.':
51                 continue
52             else:
53                 return False
54         return True

当然help function可以使用hash table的方式来检查当然元素是不是已经存在。

 1     def isValidString(self, numbers):
 2         isused = {}
 3         for elem in numbers:
 4             if elem == '.':
 5                 continue
 6             elif elem not in isused:
 7                 isused[elem] = True
 8             else:
 9                 return False
10         return True

 

posted @ 2017-02-28 07:24  lettuan  阅读(110)  评论(0)  编辑  收藏  举报