Determine if a Sudoku is valid, according to: .

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

A partially filled sudoku which is valid.

 

Note:

A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

没啥算法，纯模拟！

1 class Solution { 2 public: 3     bool isValidCell(vector
     
      
        > &board, int a, int b) { 4         vector
       
         flag(9, false); 5         int idx; 6         for (int i = 0; i < 3; ++i) { 7             for (int j = 0; j < 3; ++j) { 8                 idx = board[a + i][b + j] - '0'; 9                 if (idx > 0 && idx <= 9 && !flag[idx])  10                     flag[idx] = true;11                 else if (idx > 0 && idx <= 9 && flag[idx])12                     return false;13             }14         }15         return true;16     }17     18     bool isValidRow(vector
        
         
           > &board, int a) {19         vector
          
            flag(9, false);20 int idx;21 for (int j = 0; j < 9; ++j) {22 idx = board[a][j] - '0';23 if (idx > 0 && idx <= 9 && !flag[idx]) 24 flag[idx] = true;25 else if (idx > 0 && idx <= 9 && flag[idx])26 return false;27 }28 return true;29 }30 31 bool isValidCol(vector
           
            
              > &board, int b) {32 vector
             
               flag(9, false);33 int idx;34 for (int i = 0; i < 9; ++i) {35 idx = board[i][b] - '0';36 if (idx > 0 && idx <= 9 && !flag[idx]) 37 flag[idx] = true;38 else if (idx > 0 && idx <= 9 && flag[idx])39 return false;40 }41 return true;42 }43 44 bool isValidSudoku(vector
              
               
                 > &board) {45 for (int i = 0; i < 3; ++i) {46 for (int j = 0; j < 3; ++j) {47 if (!isValidCell(board, 3 * i, 3 * j)) 48 return false;49 }50 }51 for (int i = 0; i < 9; ++i) {52 if (!isValidRow(board, i))53 return false;54 }55 for (int j = 0; j < 9; ++j) {56 if (!isValidCol(board, j))57 return false;58 }59 return true;60 }61 };