Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy all of the following rules:
- Each of the digits
1-9
must occur exactly once in each row. - Each of the digits
1-9
must occur exactly once in each column. - Each of the digits
1-9
must occur exactly once in each of the 93x3
sub-boxes of the grid.
The '.'
character indicates empty cells.
Example 1:
Input: board = [["5","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".",".","."],[".","9","8",".",".",".",".","6","."],["8",".",".",".","6",".",".",".","3"],["4",".",".","8",".","3",".",".","1"],["7",".",".",".","2",".",".",".","6"],[".","6",".",".",".",".","2","8","."],[".",".",".","4","1","9",".",".","5"],[".",".",".",".","8",".",".","7","9"]] Output: [["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]] Explanation: The input board is shown above and the only valid solution is shown below:
Constraints:
board.length == 9
board[i].length == 9
board[i][j]
is a digit or'.'
.- It is guaranteed that the input board has only one solution.
class Solution:
def solveSudoku(self, board: List[List[str]]) -> None:
def dfs(k):
nonlocal ok
if k == len(t):
ok = True
return
i, j = t[k]
for v in range(9):
if row[i][v] == col[j][v] == block[i // 3][j // 3][v] == False:
row[i][v] = col[j][v] = block[i // 3][j // 3][v] = True
board[i][j] = str(v + 1)
dfs(k + 1)
row[i][v] = col[j][v] = block[i // 3][j // 3][v] = False
if ok:
return
row = [[False] * 9 for _ in range(9)]
col = [[False] * 9 for _ in range(9)]
block = [[[False] * 9 for _ in range(3)] for _ in range(3)]
t = []
ok = False
for i in range(9):
for j in range(9):
if board[i][j] == '.':
t.append((i, j))
else:
v = int(board[i][j]) - 1
row[i][v] = col[j][v] = block[i // 3][j // 3][v] = True
dfs(0)
class Solution {
private boolean ok;
private char[][] board;
private List<Integer> t = new ArrayList<>();
private boolean[][] row = new boolean[9][9];
private boolean[][] col = new boolean[9][9];
private boolean[][][] block = new boolean[3][3][9];
public void solveSudoku(char[][] board) {
this.board = board;
for (int i = 0; i < 9; ++i) {
for (int j = 0; j < 9; ++j) {
if (board[i][j] == '.') {
t.add(i * 9 + j);
} else {
int v = board[i][j] - '1';
row[i][v] = col[j][v] = block[i / 3][j / 3][v] = true;
}
}
}
dfs(0);
}
private void dfs(int k) {
if (k == t.size()) {
ok = true;
return;
}
int i = t.get(k) / 9, j = t.get(k) % 9;
for (int v = 0; v < 9; ++v) {
if (!row[i][v] && !col[j][v] && !block[i / 3][j / 3][v]) {
row[i][v] = col[j][v] = block[i / 3][j / 3][v] = true;
board[i][j] = (char) (v + '1');
dfs(k + 1);
row[i][v] = col[j][v] = block[i / 3][j / 3][v] = false;
}
if (ok) {
return;
}
}
}
}
using pii = pair<int, int>;
class Solution {
public:
void solveSudoku(vector<vector<char>>& board) {
bool row[9][9] = {false};
bool col[9][9] = {false};
bool block[3][3][9] = {false};
bool ok = false;
vector<pii> t;
for (int i = 0; i < 9; ++i) {
for (int j = 0; j < 9; ++j) {
if (board[i][j] == '.') {
t.push_back({i, j});
} else {
int v = board[i][j] - '1';
row[i][v] = col[j][v] = block[i / 3][j / 3][v] = true;
}
}
}
function<void(int k)> dfs = [&](int k) {
if (k == t.size()) {
ok = true;
return;
}
int i = t[k].first, j = t[k].second;
for (int v = 0; v < 9; ++v) {
if (!row[i][v] && !col[j][v] && !block[i / 3][j / 3][v]) {
row[i][v] = col[j][v] = block[i / 3][j / 3][v] = true;
board[i][j] = v + '1';
dfs(k + 1);
row[i][v] = col[j][v] = block[i / 3][j / 3][v] = false;
}
if (ok) {
return;
}
}
};
dfs(0);
}
};
func solveSudoku(board [][]byte) {
var row, col [9][9]bool
var block [3][3][9]bool
var t [][2]int
ok := false
for i := 0; i < 9; i++ {
for j := 0; j < 9; j++ {
if board[i][j] == '.' {
t = append(t, [2]int{i, j})
} else {
v := int(board[i][j] - '1')
row[i][v], col[j][v], block[i/3][j/3][v] = true, true, true
}
}
}
var dfs func(int)
dfs = func(k int) {
if k == len(t) {
ok = true
return
}
i, j := t[k][0], t[k][1]
for v := 0; v < 9; v++ {
if !row[i][v] && !col[j][v] && !block[i/3][j/3][v] {
row[i][v], col[j][v], block[i/3][j/3][v] = true, true, true
board[i][j] = byte(v + '1')
dfs(k + 1)
row[i][v], col[j][v], block[i/3][j/3][v] = false, false, false
}
if ok {
return
}
}
}
dfs(0)
}