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.

简单题，思路也非常直接，brute-force,按行扫，按列扫，按小方块扫，值得注意的是判断是否有重复元素，一个非常好的做法是使用set，一个是可以把所有不为'.'的元素加入set，判断最后set的大小是否等于不为0的元素数目。另外一个是把set当普通集合使用。但是因为python的set使用dictionary实现，判断元素在不在set中的平均时间复杂度为O(1).下面给出一个统一接口的实现，即将9个3*3的扫描做了转化，不需要3*3的循环。代码如下：

class Solution(object):
    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        for i in xrange(9):
            if not self.is_partially_valid(board,i,i+1,0,9): return False
            if not self.is_partially_valid(board,0,9,i,i+1): return False
            x = i/3*3
            y = i%3*3
            if not self.is_partially_valid(board,x,x+3,y,y+3): return False
            
        return True

    
    def is_partially_valid(self,board,x1,x2,y1,y2):
        a = set()
        for i in xrange(x1,x2):
            for j in xrange(y1,y2):
              val = board[i][j]
              if val != '.':
                  if val not in a:
                      a.add(val)
                  else:
                      return False
        return True

另外一个使用bool数字记录元素是否出现的解法也很巧妙,但是中间涉及到str到int的类型转换，代码如下：

class Solution(object):
    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        
        for i in xrange(9):
           res = [False]*9
           for j in xrange(9):
               if board[i][j]!='.':
                  if res[int(board[i][j])-1]:
                     return False
                  else:
                     res[int(board[i][j])-1] = True

        for j in xrange(9):
            res = [False]*9
            for i in xrange(9):
               if board[i][j]!='.':
                   if res[int(board[i][j])-1]:
                      return False
                   else:
                      res[int(board[i][j])-1] = True
        for i in xrange(9):
          res = [False]*9
          for j in range(i/3*3,i/3*3+3):
              for k in range(i%3*3,i%3*3+3):
                  
                  if board[j][k] !='.':
                      if res[int(board[j][k])-1]:
                        return False
                      else:
                        res[int(board[j][k])-1] = True
        return True