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='123456789'
        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