[Leetcode]36. Valid Sudoku@python

题目

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.

题目要求

判断一个数独的例子是否满足数独的要求：
1. 每行出现数字1－9一次且仅一次
2. 每列出现数字1－9一次切仅一次
3. 上图示例中粗线划出的每三行三列的block中出现数字1－9一次且仅以此

解题思路

1. 参考南郭子綦.
2. 自己实现的检验方法，看起来没有那么优美

代码

• 解法1

class Solution(object):
    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        def isValid(x,y,tmp):
            for i in range(9):
                if board[i][y] == tmp: return False
            for j in range(9):
                if board[x][j] == tmp: return False

            for i in range(3):
                for j in range(3):
                    if board[(x/3)*3 + i][(y/3)*3 + j] == tmp:
                        return False
            return True

        for i in range(9):
            for j in range(9):
                if board[i][j] == '.': continue
                tmp = board[i][j]
                board[i][j] = 'D'
                if isValid(i,j,tmp) == False: return False
                board[i][j] = tmp
        return True

• 解法2

class Solution(object):
    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        for i in range(9):
            rowMap = {}
            colMap = {}

            for j in range(9):
                # check if row valid
                if board[i][j] not in rowMap:
                    rowMap[board[i][j]] = 1
                else:
                    rowMap[board[i][j]] += 1
                    if rowMap[board[i][j]] > 1 and board[i][j] != '.':
                        return False
                # check if col valid
                if board[j][i] not in colMap:
                    colMap[board[j][i]] = 1
                else:
                    colMap[board[j][i]] += 1
                    if colMap[board[j][i]] > 1 and board[j][i] != '.':
                        return False
                # check if block valid
                if i % 3 == 0 and j % 3 == 0:
                    blockMap = {}
                    for s in range(3):
                        for t in range(3):
                            if board[i + s][j + t] not in blockMap:
                                blockMap[board[i+s][j+t]] = 1
                            else:
                                blockMap[board[i+s][j+t]] += 1
                                if blockMap[board[i+s][j+t]] > 1 and board[i+s][j+t] != '.':
                                    return False
        return True