"Leetcode" 36-valid Sudoku

Determine if a Sudoku is valid, according To:sudoku puzzles-the Rules. (http://sudoku.com.au/TheRules.aspx)

The Sudoku board could be partially filled, where empty cells is 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.

Solution 1:

Check each row, each column, and each of the nine Gongge to see if duplicate elements appear with a map record with a key unique attribute. The difficulty lies in the inspection of the ninth Gongge

Observe the law of line numbers:

No. 0 Nine Gongge: 000111222; 1th Nine Gongge: 000111222; 2nd nine Gongge: 000111222;

3rd nine Gongge: 333444555; 4th nine Gongge: 333444555; 5th nine Gongge: 333444555;

6th nine Gongge: 666777888; 7th nine Gongge: 666777888; 8th nine Gongge: 666777888;

Visible for each of the three nine Gonge increase 3, for a single nine Gongge, every three lattice line number 1.

Therefore, the row number of the J-point of the ninth Gongge can be expressed as a I/3*3+J/3

Observe the rule of the column number:

No. 0 Nine Gongge: 012012012; 1th Nine Gongge: 345345345; 2nd nine Gongge: 678678678;

3rd nine Gongge: 012012012; 4th nine Gongge: 345345345; 5th nine Gongge: 678678678;

6th nine Gongge: 012012012; 7th nine Gongge: 345345345; 8th nine Gongge: 678678678;

Visible for the next nine Gongrie increase 3, the cycle period is 3, for a single nine Gongge, each lattice point line number 1, the period is 3.

The mathematical representation of a cycle is the modulo operation MoD.

Therefore, the column number of the J-point of the ninth Gongge can be expressed as a i%3*3+j%3

1 classSolution {2  Public:3     BOOLIsvalidsudoku (vector<vector<Char>>&Board) {4           for(inti =0; I <9; i + +)5         {6unordered_map<Char,BOOL>Row; 7unordered_map<Char,BOOL>column; 8unordered_map<Char,BOOL>bod; 9              for(intj =0; J <9; J + +)Ten             { One                 if(Board[i][j]! ='.') A                 { -                     if(Row[board[i][j]] = =true)//if ROW[BOARD[I][J]] does not exist, returns 0 -                         return false; theROW[BOARD[I][J]] =true; -                 } -                 if(Board[j][i]! ='.') -                 { +                     if(Column[board[j][i]] = =true) -                         return false; +Column[board[j][i]] =true; A                 } at                 if(board[i/3*3+j/3][i%3*3+j%3] !='.') -                 { -                     if(bod[board[i/3*3+j/3][i%3*3+j%3]] ==true) -                         return false; -bod[board[i/3*3+j/3][i%3*3+j%3]] =true; -                 } in             } -         } to         return true; +     } -};

Solution 2: The solution to the cock dick seen on the discussion

1 classSolution2 {3  Public:4     BOOLIsvalidsudoku (vector<vector<Char> > &Board)5     {6         intused1[9][9] = {0}, used2[9][9] = {0}, used3[9][9] = {0};7 8          for(inti =0; I < board.size (); ++i)9              for(intj =0; J < Board[i].size (); ++j)Ten                 if(Board[i][j]! ='.') One                 { A                     intnum = board[i][j]-'0'-1, k = I/3*3+ J/3; -                     if(Used1[i][num] | | used2[j][num] | |Used3[k][num]) -                         return false; theUsed1[i][num] = Used2[j][num] = Used3[k][num] =1; -                 } -  -         return true; +     } -};

"Leetcode" 36-valid Sudoku