[Lettcode]49. Valid Sudoku Effective 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.

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

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Solution 1: Judging each row, each column, each 3x3 block has the same number, time complexity O (n^2), Spatial complexity O (n).

classSolution { Public:    BOOLIsvalidsudoku (vector<vector<Char>>&Board) {        if(Board.empty () | | board[0].empty ())return false; intn =board.size (); Vector<int> Row (9,0); Vector<int> col (9,0);  for(inti =0; I < n; ++i) {row.assign (9,0); Col.assign (9,0);  for(intj =0; J < N; ++j) {if(Board[i][j]! ='.')                {                    if(Row[board[i][j]-'1'] ==1)return false; Else++ROW[BOARD[I][J]-'1']; }                if(Board[j][i]! ='.')                {                    if(Col[board[j][i]-'1'] ==1)return false; Else++col[board[j][i]-'1']; }}} vector<int> C3 (9,0);  for(intK =0; K < n * N/9; ++k) {c3.assign (9,0); intStarti = k/3*3, STARTJ = k%3*3;  for(intcol = Starti; Col < Starti +3; ++Col) {                 for(introw = STARTJ; Row < STARTJ +3; ++row) {                    if(Board[col][row]! ='.')                    {                        if(C3[board[col][row]-'1'] ==1)return false; Else++c3[board[col][row]-'1']; }                }            }        }        return true; }};

When the spatial complexity is O (n^2), it can be solved with a two-layer loop:

classSolution { Public:    BOOLIsvalidsudoku (vector<vector<Char>>&Board) {        if(Board.empty () | | board[0].empty ())return false; intn =board.size (); Vector<int> Row (9,0); Vector<int> col (9,0); Vector< vector<int> > Cub (9, vector<int> (9,0));  for(inti =0; I < n; ++i) {row.assign (9,0); Col.assign (9,0);  for(intj =0; J < N; ++j) {if(Board[i][j]! ='.')                {                    if(Row[board[i][j]-'1'] ==1)return false; Else++ROW[BOARD[I][J]-'1']; intK = I/3*3+ J/3; if(Cub[k][board[i][j]-'1'] ==1)return false; Else++CUB[K][BOARD[I][J]-'1']; }                if(Board[j][i]! ='.')                {                    if(Col[board[j][i]-'1'] ==1)return false; Else++col[board[j][i]-'1']; }            }        }        return true; }};

[Lettcode]49. Valid Sudoku Effective Sudoku