Last Updated on 2022-11-23 by Clay
題目
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 digits1-9
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:
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 digit1-9
or'.'
.
本題是驗證一個數獨題目的有效性。老實說如果要驗證數獨題目的有效性應該再加上至少要有 17 個已填充的數值才能保證唯一解... 不過解題就是解題,題目的設定說了算。
基本上就是無論縱橫、九宮格內,數值一定都必須能夠填入 1-9。這意味著只要在發現有數值重複出現了,就代表這是一個無效的數獨題目。
解題思路
我使用的是比較很直接的解法。分別設定三個函數,在每次碰到數字時,分別判斷行、列、九宮格中是否出現同樣的數字(當然得排除自己)。
C++ 範例程式碼
class Solution {
public:
bool rowCollision(vector<vector<char>>& board, int i, int j) {
for (int k=0; k<board.size(); ++k) {
if (k != j && board[i][k] == board[i][j]) return true;
}
return false;
}
bool colCollision(vector<vector<char>>& board, int i, int j) {
for (int k=0; k<board.size(); ++k) {
if (k !=i && board[k][j] == board[i][j]) return true;
}
return false;
}
bool gridCollision(vector<vector<char>>& board, int i, int j) {
int si = i / 3 * 3;
int ei = i / 3 * 3 + 3;
int sj = j / 3 * 3;
int ej = j / 3 * 3 + 3;
for (int temp_i=si; temp_i<ei; ++temp_i) {
for (int temp_j=sj; temp_j<ej; ++temp_j) {
if (temp_i == i && temp_j == j) continue;
if (board[temp_i][temp_j] == board[i][j]) return true;
}
}
return false;
}
bool isValidSudoku(vector<vector<char>>& board) {
for (int i=0; i<board.size(); ++i) {
for (int j=0; j<board.size(); ++j) {
if (board[i][j] == '.') continue;
if (rowCollision(board, i, j) ||
colCollision(board, i, j) ||
gridCollision(board, i, j)) {
return false;
}
}
}
return true;
}
};
Python 範例程式碼
class Solution:
def rowCollision(self, board: List[List[str]], i: int, j: int) -> bool:
for k in range(len(board)):
if k != j and board[i][k] == board[i][j]:
return True
return False
def colCollision(self, board: List[List[str]], i: int, j: int) -> bool:
for k in range(len(board)):
if k != i and board[k][j] == board[i][j]:
return True
return False
def gridCollision(self, board: List[List[str]], i: int, j: int) -> bool:
si = i // 3 * 3
ei = i // 3 * 3 + 3
sj = j // 3 * 3
ej = j // 3 * 3 + 3
for temp_i in range(si, ei):
for temp_j in range(sj, ej):
if temp_i == i and temp_j == j:
continue
if board[temp_i][temp_j] == board[i][j]:
return True
return False
def isValidSudoku(self, board: List[List[str]]) -> bool:
for i in range(len(board)):
for j in range(len(board)):
if board[i][j] == ".":
continue
if self.rowCollision(board, i, j) or self.colCollision(board, i, j) or self.gridCollision(board, i, j):
return False
return True