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.
思考:参考这里。
class Solution { public: // 返回第一个空白的位置,如果没找到就返回 (-1, -1) pair<int, int> findFirstEmpty(const vector< vector<char> >& board) { for (int i = 0; i < 9; ++i) for (int j = 0; j < 9; ++j) if (board[i][j] == '.') return make_pair(i, j); return make_pair(-1, -1); } // 检查连续的 9 个格子是否有效 bool isValid(const vector<char>& vec) { vector<bool> occur(9, false); for (int i = 0; i < 9; ++i) { if (isdigit(vec[i])) { if (occur[vec[i]-'1']) return false; else occur[vec[i]-'1'] = true; } } return true; } // 检查往某个位置填入一个数之后整个 board 是否有效(只需要考虑当前行、 // 当前列和所属的田字格) bool isValidBoard(const vector< vector<char> >& board, pair<int, int> pos) { // 检查当前行是否有效 if (!isValid(board[pos.first])) return false; // 检查当前列是否有效 vector<char> column(9); for (int i = 0; i < 9; ++i) column[i] = board[i][pos.second]; if (!isValid(column)) return false; // 检查所在的田字格是否有效 int block_row = pos.first / 3; int block_col = pos.second / 3; vector<char> block; for (int i = block_row * 3; i < block_row * 3 + 3; ++i) for (int j = block_col * 3; j < block_col * 3 + 3; ++j) block.push_back(board[i][j]); if (!isValid(block)) return false; // 如果以上都有效,则返回 true return true; } // 检查从当前局面开始是否能够得到最终合法有效的解 bool solveSudoku(vector<vector<char> >& board) { // 如果没有找到空白的格子,说明已经填满了,成功返回 pair<int, int> pos = findFirstEmpty(board); if (pos.first == -1 && pos.second == -1) return true; // 否则依次尝试往当前格子中填入数字 1-9,并判断能否得到可行的解 for (int i = 1; i <= 9; ++i) { board[pos.first][pos.second] = i + '0'; if (isValidBoard(board, pos) && solveSudoku(board)) return true; // 恢复原样 board[pos.first][pos.second] = '.'; } return false; } };