[leetcode] 36. Valid Sudoku @ python

原题

https://leetcode.com/problems/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 1-9 without repetition.
Each column must contain the digits 1-9 without repetition.
Each of the 9 3x3 sub-boxes of the grid must contain the digits 1-9 without 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 1-9 and the character ‘.’.
The given board size is always 9x9.

解法

使用分治法, 分别检查行, 列, 小格, 将读取的列表转化为字典, 检查字典是否有重复的数字.
Time: 2*O(n) + O(n**2/9)
Space: O(1)

代码

class Solution(object):
    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        def check(count):
            for char in count:
                if char != '.' and count[char] > 1:
                    return False
            return True
    
        # check row
        for row in board:
            count_row = collections.Counter(row)
            if check(count_row) == False:
                return False
        # check col
        n = len(board)
        for c in range(n):
            count_col = collections.Counter([board[r][c] for r in range(n)])
            if check(count_col) == False:
                return False
        # check sub boxes
        for x in range(0, n, 3):
            for y in range(0, n, 3):
                sub = [board[x+dx][y+dy] for dx in range(3) for dy in range(3)]
                count_sub = collections.Counter(sub)
                if check(count_sub) == False:
                    return False
            
        return True  
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