36. Valid Sudoku (Array; HashTable)
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.
class Solution {
public:
bool isValidSudoku(vector<vector<char> > &board) {
vector<bool> flag(, false); //indicate the appearance of 1,2,..., 9
//check line
for(int i = ; i<; i++)
{
for(int j = ; j<; j++)
{
if(board[i][j]=='.') continue;
if(flag[board[i][j]-'']) return false;
flag[board[i][j]-''] = true;
}
flag.assign(board.size(),false); //reset the vector
} //check column
for(int j = ; j<; j++)
{
for(int i = ; i<; i++)
{
if(board[i][j]=='.') continue;
if(flag[board[i][j]-'']) return false;
flag[board[i][j]-''] = true;
}
flag.assign(board.size(),false);
} //check small square
for(int i = ; i < ; i+=){
for(int j = ; j <; j+=){
for(int m = ; m < ; m++){
for(int n = ; n < ; n++){
if(board[i+m][j+n]=='.') continue;
if(flag[board[i+m][j+n]-'']) return false;
flag[board[i+m][j+n]-''] = true;
}
}
flag.assign(board.size(),false);
}
}
return true;
}
};
如果没有'.',那么我们可以用以下的方法判断是否valid (参数是某一行,某一列,或是某一个small square的9个元素)
bool check(vector<int> v) {
sort(v.begin(), v.end());
for (int i = ; i < (int)v.size(); ++i) {
if (v[i] != i + ) {
return false;
}
}
return true;
}
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