leetcode - 36.Valid Sudoku

Valid Sudoku

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.

Solution1:

public boolean isValidSudoku(char[][] board) {
// -
for (char[] chars : board) {
Set<Character> set = new HashSet<>();
for (char c : chars) {
if (!canAdd(set, c)) {
return false;
}
}
}

// |
int i = 0;
while (i < 9) {
Set<Character> set = new HashSet<>();
for (char[] chars : board) {
char c = chars[i];
if (!canAdd(set, c)) {
return false;
}
}
i++;
}

// ---
// | |
// ---
return isValid(board, 0, 1, 2, 0, 1, 2) && isValid(board, 3, 4, 5, 0, 1, 2) && isValid(board, 6, 7, 8, 0, 1, 2) //
&& isValid(board, 0, 1, 2, 3, 4, 5) && isValid(board, 3, 4, 5, 3, 4, 5)
&& isValid(board, 6, 7, 8, 3, 4, 5) //
&& isValid(board, 0, 1, 2, 6, 7, 8) && isValid(board, 3, 4, 5, 6, 7, 8)
&& isValid(board, 6, 7, 8, 6, 7, 8);
}

private boolean isValid(char[][] board, int x1, int x2, int x3, int y1, int y2, int y3) {
Set<Character> set = new HashSet<>();
return canAdd(set, board[x1][y1]) && canAdd(set, board[x2][y1]) && canAdd(set, board[x3][y1]) //
&& canAdd(set, board[x1][y2]) && canAdd(set, board[x2][y2]) && canAdd(set, board[x3][y2]) //
&& canAdd(set, board[x1][y3]) && canAdd(set, board[x2][y3]) && canAdd(set, board[x3][y3]);
}

private boolean canAdd(Set<Character> set, char c) {
if (c == '.' || set.add(c)) {
return true;
}
return false;
}

Solution2:

better

public boolean isValidSudoku1(char[][] board) {
for(int i = 0; i < 9; i++){
HashSet<Character> row = new HashSet<>();
HashSet<Character> column = new HashSet<>();
HashSet<Character> cube = new HashSet<>();
for(int j = 0; j < 9; j++){
if(board[i][j] != '.' && !row.add(board[i][j])){
return false;
}
if(board[j][i] != '.' && !column.add(board[j][i])){
return false;
}
int rowIndex = (i / 3) * 3 + j / 3;
int columnIndex = (i % 3) * 3 + j % 3;
if(board[rowIndex][columnIndex] != '.' && !cube.add(board[rowIndex][columnIndex])){
return false;
}
}
}
return true;
}