Write a program to solve a Sudoku puzzle by filling the empty cells.
Empty cells are indicated by the character '.'.
You may assume that there will be only one unique solution.
A sudoku puzzle...
...and its solution numbers marked in red.
这道求解数独的题是在之前那道的基础上的延伸，之前那道题让我们验证给定的数组是否为数独数组，这道让我们求解数独数组，跟此题类似的有，，等等，其中尤其是跟的解题思路及其相似，对于每个需要填数字的格子带入1到9，每代入一个数字都判定其是否合法，如果合法就继续下一次递归，结束时把数字设回'.'，判断新加入的数字是否合法时，只需要判定当前数字是否合法，不需要判定这个数组是否为数独数组，因为之前加进的数字都是合法的，这样可以使程序更加高效一些，具体实现如代码所示：
class Solution {public: void solveSudoku(vector > &board) { if (board.empty() || board.size() != 9 || board[0].size() != 9) return; solveSudokuDFS(board, 0, 0); } bool solveSudokuDFS(vector > &board, int i, int j) { if (i == 9) return true; if (j >= 9) return solveSudokuDFS(board, i + 1, 0); if (board[i][j] == '.') { for (int k = 1; k <= 9; ++k) { board[i][j] = (char)(k + '0'); if (isValid(board, i , j)) { if (solveSudokuDFS(board, i, j + 1)) return true; } board[i][j] = '.'; } } else { return solveSudokuDFS(board, i, j + 1); } return false; } bool isValid(vector > &board, int i, int j) { for (int col = 0; col < 9; ++col) { if (col != j && board[i][j] == board[i][col]) return false; } for (int row = 0; row < 9; ++row) { if (row != i && board[i][j] == board[row][j]) return false; } for (int row = i / 3 * 3; row < i / 3 * 3 + 3; ++row) { for (int col = j / 3 * 3; col < j / 3 * 3 + 3; ++col) { if ((row != i || col != j) && board[i][j] == board[row][col]) return false; } } return true; }}; 本文转自博客园Grandyang的博客，原文链接：，如需转载请自行联系原博主。