Description#
Determine if a 9 x 9
Sudoku board is valid. Only the filled cells need to be validated according to the following rules:
- Each row must contain the digits
1-9
without repetition. - Each column must contain the digits
1-9
without repetition. - Each of the nine
3 x 3
sub-boxes of the grid must contain the digits 1-9
without repetition.
Note:
- A Sudoku board (partially filled) could be valid but is not necessarily solvable.
- Only the filled cells need to be validated according to the mentioned rules.
Example 1:
![](https://spcdn.pages.dev/leetcode/problems/0036.Valid%20Sudoku/images/250px-Sudoku-by-L2G-20050714.svg.png)
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: true
Example 2:
Input: board =
[["8","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: false
Explanation: Same as Example 1, except with the 5 in the top left corner being modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.
Constraints:
board.length == 9
board[i].length == 9
board[i][j]
is a digit 1-9
or '.'
.
Solutions#
Solution 1: Traversal once#
The valid sudoku satisfies the following three conditions:
- The digits are not repeated in each row;
- The digits are not repeated in each column;
- The digits are not repeated in each $3 \times 3$ box.
Traverse the sudoku, for each digit, check whether the row, column and $3 \times 3$ box it is in have appeared the digit. If it is, return false
. If the traversal is over, return true
.
The time complexity is $O(C)$ and the space complexity is $O(C)$, where $C$ is the number of empty spaces in the sudoku. In this question, $C=81$.
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| class Solution:
def isValidSudoku(self, board: List[List[str]]) -> bool:
row = [[False] * 9 for _ in range(9)]
col = [[False] * 9 for _ in range(9)]
sub = [[False] * 9 for _ in range(9)]
for i in range(9):
for j in range(9):
c = board[i][j]
if c == '.':
continue
num = int(c) - 1
k = i // 3 * 3 + j // 3
if row[i][num] or col[j][num] or sub[k][num]:
return False
row[i][num] = True
col[j][num] = True
sub[k][num] = True
return True
|
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| class Solution {
public boolean isValidSudoku(char[][] board) {
boolean[][] row = new boolean[9][9];
boolean[][] col = new boolean[9][9];
boolean[][] sub = new boolean[9][9];
for (int i = 0; i < 9; ++i) {
for (int j = 0; j < 9; ++j) {
char c = board[i][j];
if (c == '.') {
continue;
}
int num = c - '0' - 1;
int k = i / 3 * 3 + j / 3;
if (row[i][num] || col[j][num] || sub[k][num]) {
return false;
}
row[i][num] = true;
col[j][num] = true;
sub[k][num] = true;
}
}
return true;
}
}
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| class Solution {
public:
bool isValidSudoku(vector<vector<char>>& board) {
vector<vector<bool>> row(9, vector<bool>(9, false));
vector<vector<bool>> col(9, vector<bool>(9, false));
vector<vector<bool>> sub(9, vector<bool>(9, false));
for (int i = 0; i < 9; ++i) {
for (int j = 0; j < 9; ++j) {
char c = board[i][j];
if (c == '.') continue;
int num = c - '0' - 1;
int k = i / 3 * 3 + j / 3;
if (row[i][num] || col[j][num] || sub[k][num]) {
return false;
}
row[i][num] = true;
col[j][num] = true;
sub[k][num] = true;
}
}
return true;
}
};
|
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| func isValidSudoku(board [][]byte) bool {
row, col, sub := [9][9]bool{}, [9][9]bool{}, [9][9]bool{}
for i := 0; i < 9; i++ {
for j := 0; j < 9; j++ {
num := board[i][j] - byte('1')
if num < 0 || num > 9 {
continue
}
k := i/3*3 + j/3
if row[i][num] || col[j][num] || sub[k][num] {
return false
}
row[i][num] = true
col[j][num] = true
sub[k][num] = true
}
}
return true
}
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| function isValidSudoku(board: string[][]): boolean {
const row: boolean[][] = Array.from({ length: 9 }, () =>
Array.from({ length: 9 }, () => false),
);
const col: boolean[][] = Array.from({ length: 9 }, () =>
Array.from({ length: 9 }, () => false),
);
const sub: boolean[][] = Array.from({ length: 9 }, () =>
Array.from({ length: 9 }, () => false),
);
for (let i = 0; i < 9; ++i) {
for (let j = 0; j < 9; ++j) {
const num = board[i][j].charCodeAt(0) - '1'.charCodeAt(0);
if (num < 0 || num > 8) {
continue;
}
const k = Math.floor(i / 3) * 3 + Math.floor(j / 3);
if (row[i][num] || col[j][num] || sub[k][num]) {
return false;
}
row[i][num] = true;
col[j][num] = true;
sub[k][num] = true;
}
}
return true;
}
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| /**
* @param {character[][]} board
* @return {boolean}
*/
var isValidSudoku = function (board) {
const row = [...Array(9)].map(() => Array(9).fill(false));
const col = [...Array(9)].map(() => Array(9).fill(false));
const sub = [...Array(9)].map(() => Array(9).fill(false));
for (let i = 0; i < 9; ++i) {
for (let j = 0; j < 9; ++j) {
const num = board[i][j].charCodeAt() - '1'.charCodeAt();
if (num < 0 || num > 8) {
continue;
}
const k = Math.floor(i / 3) * 3 + Math.floor(j / 3);
if (row[i][num] || col[j][num] || sub[k][num]) {
return false;
}
row[i][num] = true;
col[j][num] = true;
sub[k][num] = true;
}
}
return true;
};
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