# [Leetcode] Sudoku Solver @ Python

Source: Internet
Author: User
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Test instructions

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

Empty cells is indicated by the character ‘.‘ .

Assume that there would be is only one unique solution.

A Sudoku Puzzle ...

... and its solution numbers marked in red.

Problem Solving Ideas 1: (This version was tested in 2014/09/12 just passed)

Find the solution of Sudoku, dancing links will not, can only write single compression version.

http://c4fun.cn/blog/2014/03/20/leetcode-solution-02/#Sudoku_Solver

classSolution:#@param board, a 9x9 2D array    #Solve the Sudoku by modifying the input board In-place.    #Don't return any value.    defSolvesudoku (self, Board): LT, RT, BT= [0] * 9, [0] * 9, [0] * 9Self.dt= {}         forIinchRange (9): self.dt[1<<i] = Chr (ORD ('1')+i) forIinchRange (9): Board[i]=list (Board[i]) forJinchRange (9):                if(Board[i][j] = ='.'):                    Continue; Num= Ord (Board[i][j])-Ord ('1') Lt[i]|= 1 <<Num RT[J]|= 1 <<Num bt[j3*3+I//3] |= 1 <<Num SELF.DFS (board, 0, LT, RT, BT) Board= ["'. Join (s) forSinchboard]defDfs (self, board, p, LT, RT, BT): whileP < 81 andBOARD[P/9][P%9]! ='.': P+ = 1ifp = = 81:            returnTrue I, J, K= P//9, p%9, P%9//3*3+P//9//3ifBOARD[I][J]! ='.': Self.dfs (board, p+ 1, LT, RT, BT)returnTrue can= (~ (Lt[i]|rt[j]|bt[k])) & (0x1ff) Pre=Board[i] whileCan:num= can&-can board[i][j]=Self.dt[num] Lt[i]|=Num RT[J]|=Num Bt[k]|=NumifSelf.dfs (board, p + 1, LT, RT, BT):returnTrue Board[i][j]='.'Lt[i]&= ~Num RT[J]&= ~Num Bt[k]&= ~Num can-=NumreturnFalse

Problem Solving idea 2: (this version was tested in 2014/09/12, timed out, failed)

Use DFS to resolve the issue.

Http://www.cnblogs.com/zuoyuan/p/3770271.html

classSolution:#@param board, a 9x9 2D array    #Solve the Sudoku by modifying the input board In-place.    #Don't return any value.    defSolvesudoku (self, Board):defisValid (x, y): tmp=board[x][y]; board[x][y]='D'             forIinchRange (9):                ifBoard[i][y]==tmp:returnFalse forIinchRange (9):                ifBoard[x][i]==tmp:returnFalse forIinchRange (3):                 forJinchRange (3):                    ifboard[(X/3) *3+i][(Y/3) *3+j]==tmp:returnFalse Board[x][y]=tmpreturnTruedefDFS (board): forIinchRange (9):                 forJinchRange (9):                    ifboard[i][j]=='.':                         forKinch '123456789': Board[i][j]=kifIsValid (I,J) andDFS (board):returnTrue Board[i][j]='.'                        returnFalsereturnTrue DFS (board)

[Leetcode] Sudoku Solver @ Python

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