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[LeetCode] Valid Sudoku
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; }
[LeetCode] Valid Sudoku
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