Judging whether Sudoku is legal (Lintcode)

Judging whether Sudoku is legalPlease determine if a Sudoku is valid.

The Sudoku may only be populated with partial numbers, where missing numbers are . represented.

Sample Example

The following is an example of a legal sudoku.

Note

A valid Sudoku (partial padding only) is not necessarily solvable. We just need to make the padding space valid.

Description

What's that 数独 ?

• Http://sudoku.com.au/TheRules.aspx
• Http://baike.baidu.com/subview/961/10842669.htm

The idea of a time-out in the first place would be to have a space for a moment. The following code appears:

After a thought, should not time out to put. But this code is actually a lot simpler.

Total time: 1005 ms

1 classSolution {2     /**3       * @paramboard:the Board4         @return: Wether the Sudoku is valid5       */6      Public BooleanIsvalidsudoku (Char[] board) {7         intR[][] =New int[9] [9];8         intL[][] =New int[9] [9];9         intA[][][] =New int[3] [3] [9];Ten          One          for(inti=0;i<9;i++) { A              for(intj=0;j<9;j++) { -R[I][J] = 0; -L[I][J] = 0; the                 if(I < 3 && J < 3) { -                      for(intk = 0; k<9; k++) -A[i][j][k] = 0; -                 } +             } -         } +          A          for(inti=0;i<9;i++) { at              for(intj=0;j<9;j++) { -                 if(Board[i][j] = = '. ')Continue; -                 intn = board[i][j]-' 0 '; -                 if(N < 0 | | n > 9)return false; -                 if(++r[i][n-1] > 1)return false; -                 if(++l[j][n-1] > 1)return false; in                 intx = I/3; -                 inty = J/3; to                 if(++a[x][y][n-1] > 1)return false; +             } -         } the          *         return true; $     }Panax Notoginseng};

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Judging whether Sudoku is legal (Lintcode)