Valid Sudoku

题目

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.

方法

只需要判断每一行，每一列以及每一个小矩形里面没有重复的元素即可。如果是空着的，默认为正确的。

private boolean isValidRow(char[][] board) {
int len = 9;
boolean[] status = new boolean[len];
for (int i = 0; i < len; i++) {
for (int k = 0; k < len; k++) {
status[k] = false;
}

for (int j = 0; j < len; j++) {
if (board[i][j] != '.') {
if (status[board[i][j] - '1']) {
return false;
} else {
status[board[i][j] - '1'] = true;
}
}
}

}
return true;
}

private boolean isValidColumn(char[][] board) {
int len = 9;
boolean[] status = new boolean[len];
for (int i = 0; i < len; i++) {
for (int k = 0; k < len; k++) {
status[k] = false;
}

for (int j = 0; j < len; j++) {
if (board[j][i] != '.') {
if (status[board[j][i] - '1']) {
return false;
} else {
status[board[j][i] - '1'] = true;
}
}
}
}
return true;
}

private boolean isValidBoxes(char[][] board) {
int len = 9;
boolean[] status = new boolean[len];
for (int i = 0; i < len; i = i + 3) {
for (int j = 0; j < len; j = j + 3) {
for (int k = 0; k < len; k++) {
status[k] = false;
}

for (int p = 0; p < 3; p++) {
for (int q = 0; q < 3; q++) {
if (board[i + p][j + q] != '.') {
if (status[board[i + p][j + q] - '1']) {
return false;
} else {
status[board[i + p][j + q] - '1'] = true;
}
}
}
}

}
}

return true;
}
public boolean isValidSudoku(char[][] board) {
return isValidRow(board) && isValidColumn(board) && isValidBoxes(board);
}