Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.
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.
Analysis:
first scan every row and column. then scan every square.
Java
public boolean isValidSudoku(char[][] board) {
ArrayList<Character> row = new ArrayList<>();
ArrayList<Character> col = new ArrayList<>();
for(int i=0;i<9;i++){
row.clear();
col.clear();
for(int j=0;j<9;j++){
if(board[i][j]!='.'){
if(col.contains(board[i][j]))
return false;
else
col.add(board[i][j]);
}
if(board[j][i]!='.'){
if(row.contains(board[j][i]))
return false;
else
row.add(board[j][i]);
}
}
}
ArrayList<Character> block = new ArrayList<>();
for(int i=0;i<9;i=i+3){
for(int j=0;j<9;j=j+3){
block.clear();
for(int m=0;m<3;m++){
for(int n = 0;n<3;n++){
if(board[i+m][j+n]!='.')
if(block.contains(board[i+m][j+n])) return false;
else block.add(board[i+m][j+n]);
}
}
}
}
return true;
}
c++
bool isValidSudoku(vector<vector<char> > &board) {
for(int i=0; i<board.size(); i++){
for(int j=0; j<board[i].size();j++){
for(int k=j+1; k<board[i].size();k++){
if(board[i][j] == board[i][k] && board[i][j]!='.')
return false;
}
for(int k=0; k<board.size(); k++){
if(board[i][j] == board[k][j] && board[i][j]!='.' && i!=k)
return false;
}
}
}
int row[9];
for(int i=0; i<9;i+=3){
for(int j=0; j<9; j+=3){
memset(row,0,9*sizeof(int));
for(int m=0; m<3;m++){
for(int n=0; n<3;n++){
if(board[i+m][j+n] == '.')
continue;
if(row[board[m+i][j+n]-49] == 1)
return false;
row[board[m+i][j+n]-49]++;
}
}
}
}
return true;
}