[Leedcode 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 is filled with the character ‘.‘ .

A partially filled sudoku which is valid.

 Public classSolution { Public BooleanIsvalidsudoku (Char[] board) {        //The idea: first each row of each column to determine whether there is a number of repetitions, the problem with the help of ArrayList, pay attention to Clear,contains,add and other methods, when the board element is not '. ' Only when the judgment is made.//judge the rows and columns to determine the block of each 3*3. Note the way of traversal, four for loops        intlen=board.length; List<Character> row=NewArraylist<character>(); List<Character> col=NewArraylist<character>();  for(inti=0;i<len;i++) {row.clear ();            Col.clear ();  for(intj=0;j<len;j++){                                CharR=Board[i][j]; if(r!= '. ')){                    if(Row.contains (R))return false; Else{Row.add (R); }                }                CharC=Board[j][i]; if(c!= '. ')){                    if(Col.contains (c))return false; Else{Col.add (c); } }}} ArrayList<Character> block=NewArraylist<character>();  for(intK=0;K&LT;LEN;K=K+3) {//K represents the row of the block, and S represents the column of the block. Each block is then traversed             for(intS=0;s<len;s=s+3) {block.clear ();  for(inti=0;i<3;i++){                     for(intj=0;j<3;j++){                      Charb=board[i+k][j+S]; if(b!= '. ')){                            if(Block.contains (b))return false; Else{Block.add (b); }                        }                      }                }            }        }                      return true; }}

[Leedcode 36] Valid Sudoku