Problem:
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.
Idea:
DFS + backtracking
Solution:
C++
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 |
// Author: Huahua // Runtime: 3 ms class Solution { public: void solveSudoku(vector<vector<char>>& board) { rows_ = vector<vector<int>>(9, vector<int>(10)); cols_ = vector<vector<int>>(9, vector<int>(10)); boxes_ = vector<vector<int>>(9, vector<int>(10)); for(int i = 0; i < 9; i++) for(int j = 0; j < 9; j++) { const char c = board[i][j]; if ( c != '.') { int n = c - '0'; int bx = j / 3; int by = i / 3; rows_[i][n] = 1; cols_[j][n] = 1; boxes_[by * 3 + bx][n] = 1; } } fill(board, 0, 0); } private: vector<vector<int>> rows_, cols_, boxes_; bool fill(vector<vector<char>>& board, int x, int y) { if (y == 9) return true; int nx = (x + 1) % 9; int ny = (nx == 0) ? y + 1 : y; if (board[y][x] != '.') return fill(board, nx, ny); for (int i = 1; i <= 9; i++) { int bx = x / 3; int by = y / 3; int box_key = by * 3 + bx; if (!rows_[y][i] && !cols_[x][i] && !boxes_[box_key][i]) { rows_[y][i] = 1; cols_[x][i] = 1; boxes_[box_key][i] = 1; board[y][x] = i + '0'; if (fill(board, nx, ny)) return true; board[y][x] = '.'; boxes_[box_key][i] = 0; cols_[x][i] = 0; rows_[y][i] = 0; } } return false; } }; |
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