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.

思路：

思路很简单，只要检查每行、每列、每个子区域中没有重复的数字即可。

代码：

1     bool isValidSudoku(vector
     
      
        > &board) { 2         // Note: The Solution object is instantiated only once and is reused by each test case. 3         map
       
         table; 4         for(int i = 0; i < 9; i++){ 5             table.clear(); 6             for(int j = 0; j < 9; j++){ 7                 if(board[i][j] != '.'){ 8                     if(table.find(board[i][j]) != table.end()) 9                         return false;10                     else11                         table[board[i][j]] = 1;12                 }13             }14         }15         for(int i = 0; i < 9; i++){16             table.clear();17             for(int j = 0; j < 9; j++){18                 if(board[j][i] != '.'){19                     if(table.find(board[j][i]) != table.end())20                         return false;21                     else22                         table[board[j][i]] = 1;23                 }24             }25         }26         for(int i = 0; i < 9; i++){27             table.clear();28             for(int j = 0; j < 3; j++){29                 for(int k = 0; k < 3; k++){30                     if(board[3*(i/3)+j][3*(i%3)+k] != '.'){31                         if(table.find(board[3*(i/3)+j][3*(i%3)+k]) != table.end())32                             return false;33                         else34                             table[board[3*(i/3)+j][3*(i%3)+k]] = 1;35                     }36                 }37             }38         }39         return true;40     }