Leetcode||36. Valid Sudoku

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.

数独的规则：数独的每一行、每一列、9个九宫格里的数字只能是0-9，且不能重复。

经典一点，就暴力破解，三个规则都过一遍

class Solution(object):    def isValidSudoku(self, board):        """        :type board: List[List[str]]        :rtype: bool        """        for i in range(9):            if not self.isValidNine(board[i]):                return False            col = [c[i] for c in board]            if not self.isValidNine(col):                return False        for i in [0, 3, 6]:            for j in [0, 3, 6]:                block = [board[s][t] for s in [i, i+1, i+2] for t in [j, j+1, j+2]]                if not self.isValidNine(block):                    return False        return True    def isValidNine(self, row):        map = {}        for c in row:            if c != '.':                if c in map:                    return False                else:                    map[c] = True        return True

网上找了一个非常短小的代码，值得学习，方法是记录所有出现过的行、列、块的数字及相应位置，最后判断是否有重复，用set操作精简代码

class Solution(object):    def isValidSudoku(self, board):        """        :type board: List[List[str]]        :rtype: bool        """        seen = []        for i, row in enumerate(board):        for j, col in enumerate(row):        if col != '.':        seen += [(col, j), (i, col), (i/3, j/3, col)]        return len(seen) == len(set(seen))