标签：blog http ar for sp div on art log
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.
这题看似很复杂，其实很无脑，因为也无需判断是否是可解的sudoku board。遍历这个board，找到不为空的位置时，分别考察其行，列和3x3；如果出现重复则为非法，遍历完毕全部通过检查则为合法。
bool isValidSudoku(vector<vector<char> > &board) {
unordered_set<int> column;
unordered_set<int> square;
unordered_set<int> row;
// check for colum
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
char item = board[j][i];
if (item != ‘.‘) {
int num = atoi(&item);
if (column.find(num) == column.end()){
column.insert(atoi(&item));
}
else {
return false;
}
}
}
column.clear();
}
// check for row
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
char item = board[i][j];
if (item != ‘.‘) {
int num = atoi(&item);
if (row.find(num) == row.end()) {
row.insert(num);
}
else {
return false;
}
}
}
row.clear();
}
// check for 3x3
for (int offsetY = 0; offsetY < 3; offsetY++) {
for (int offsetX = 0; offsetX < 3; offsetX++) {
for (int i = 3*offsetY; i < 3*offsetY+3; i++) {
for (int j = 3*offsetX; j < 3*offsetX+3; j++) {
char item = board[i][j];
if (item != ‘.‘) {
int num = atoi(&item);
if (square.find(num) == square.end()) {
square.insert(num);
}
else {
return false;
}
}
}
}
square.clear();
}
}
return true;
}
标签：blog http ar for sp div on art log
原文地址：http://www.cnblogs.com/agentgamer/p/4072725.html
