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.
这道题让我们验证一个方阵是否为数独矩阵，判断标准是看各行各列是否有重复数字，以及每个小的3x3的小方阵里面是否有重复数字，如果都无重复，则当前矩阵是数独矩阵，但不代表待数独矩阵有解，只是单纯的判断当前未填完的矩阵是否是数独矩阵。那么根据数独矩阵的定义，我们在遍历每个数字的时候，就看看包含当前位置的行和列以及3x3小方阵中是否已经出现该数字，那么我们需要三个标志矩阵，分别记录各行，各列，各小方阵是否出现某个数字，其中行和列标志下标很好对应，就是小方阵的下标需要稍稍转换一下，具体代码如下：
class Solution { public: bool isValidSudoku(vector<vector<char> > &board) { if (board.empty() || board[0].empty()) return false; int m = board.size(), n = board[0].size(); vector<vector<bool> > rowFlag(m, vector<bool>(n, false)); vector<vector<bool> > colFlag(m, vector<bool>(n, false)); vector<vector<bool> > cellFlag(m, vector<bool>(n, false)); for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { if (board[i][j] >= '1' && board[i][j] <= '9') { int c = board[i][j] - '1'; if (rowFlag[i][c] || colFlag[c][j] || cellFlag[3 * (i / 3) + j / 3][c]) return false; rowFlag[i][c] = true; colFlag[c][j] = true; cellFlag[3 * (i / 3) + j / 3][c] = true; } } } return true; } };
本文转自博客园Grandyang的博客，原文链接：验证数独[LeetCode] Valid Sudoku ，如需转载请自行联系原博主。
