LEETCODE —— Sudoku Solver

Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells are indicated by the character '.'.

You may assume that there will be only one unique solution.

A sudoku puzzle...

...and its solution numbers marked in red.

 class Solution(object):
     def validset(self ):
         s=''
         return set(s)

     def initTbl(self, tbl, board):
         for i in range(9):
             for j in range(9):
                 if board[i][j] == '.':
                     tbl[(i,j)]=[]

     def solveSudoku(self, board):
         """
         :type board: List[List[str]]
         :rtype: bool
         """
         resultTbl={}
         visited=[]
         self.initTbl(resultTbl, board)
         reuse=False
         unvisited = resultTbl.keys()
         unvisited.sort()
         unvisited.reverse()

         while unvisited != []:
             (x, y) = unvisited.pop()
             if reuse==False:
                 resultTbl[(x,y)] = self.possibleValues((x,y), board)
                 if resultTbl[(x,y)] == None: # invalid sudoku
                     print False
                     break

             if len( resultTbl[(x,y)] ) == 0: # DEAD END, BACKTRACK
                 unvisited.append((x, y))
                 if visited != []:
                     reuse=True
                     prev=visited.pop()
                     unvisited.append(prev)
                     x=prev[0]
                     y=prev[1]
                     board[x][y]='.'
                 else:
                     break
                 continue
             board[x][y]=resultTbl[(x,y)].pop()
             visited.append((x,y))
             reuse=False
         for line in board:
             if '.' in line:
                 print False

     def possibleValues(self, coord, board):
         vals = {'.':0}
         for i in range(1,10): #init
             vals[str(i)]=0

         for y in range(0,9):
             node=board[coord[0]][y]
             vals[node]+=1
             if vals[node]>1 and node!='.': return None
         for x in range(0,9):
             node=board[x][coord[1]]
             vals[node]+=1
             if vals[node] > 2 and node!='.': return None
         x = coord[0]/3*3
         y = coord[1]/3*3
         for i in range(x, x+3):
             for j in range(y, y+3):
                 node=board[i][j]
                 vals[node]+=1
                 if vals[node]>3 and node!='.': return None
         s = set()
         for k in vals.keys():
             if vals[k]>=1: s.add(k)
         return self.validset() - s