37 Sudoku Solver 解数独
Description:
Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy all of the following rules:
Each of the digits 1-9 must occur exactly once in each row.
Each of the digits 1-9 must occur exactly once in each column.
Each of the the digits 1-9 must occur exactly once in each of the 9 3x3 sub-boxes of the grid.
Empty cells are indicated by the character '.'.
A sudoku puzzle...
...and its solution numbers marked in red.
Note:
The given board contain only digits 1-9 and the character '.'.
You may assume that the given Sudoku puzzle will have a single unique solution.
The given board size is always 9x9.
题目描述:
编写一个程序，通过已填充的空格来解决数独问题。
一个数独的解法需遵循如下规则：
数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。
空白格用 '.' 表示。
一个数独。
答案被标成红色。
说明:
给定的数独序列只包含数字 1-9 和字符 '.' 。
你可以假设给定的数独只有唯一解。
给定数独永远是 9x9 形式的。
思路:
回溯法
- 初始化条件, 在这里是给出题目中已经存在的数字
- 做选择, 从 1-9中选出满足 row, col, box的进入下一层函数选择
- 撤销选择
- 满足条件的加入结果
时间复杂度O(1), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
void solveSudoku(vector<vector<char>>& board)
{
bool row[9][9]{false}, col[9][9]{false}, box[9][9]{false};
for (int i = 0; i < 9; i++)
{
for (int j = 0; j < 9; j++)
{
int num = board[i][j] - '1';
if (num > -1)
{
row[i][num] = true;
col[j][num] = true;
box[i / 3 * 3 + j / 3][num] = true;
}
}
}
helper(board, row, col, box, 0, 0);
}
private:
bool helper(vector<vector<char>>& board, bool row[9][9], bool col[9][9], bool box[9][9], int i, int j)
{
if (j == 9)
{
j = 0;
++i;
if (i == 9) return true;
}
if (board[i][j] == '.')
{
for (int num = 0; num < 9; num++)
{
bool valid = !(row[i][num] or col[j][num] or box[i / 3 * 3 + j / 3][num]);
if (valid)
{
row[i][num] = true;
col[j][num] = true;
box[i / 3 * 3 + j / 3][num] = true;
board[i][j] = (char)(num + '1');
if (helper(board, row, col, box, i, j + 1)) return true;
board[i][j] = '.';
row[i][num] = false;
col[j][num] = false;
box[i / 3 * 3 + j / 3][num] = false;
}
}
}
else return helper(board, row, col, box, i, j + 1);
return false;
}
};
Java:
class Solution {
public void solveSudoku(char[][] board) {
boolean row[][] = new boolean[9][9], col[][] = new boolean[9][9], box[][] = new boolean[9][9];
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
int num = board[i][j] - '1';
if (num > -1) {
row[i][num] = true;
col[j][num] = true;
box[i / 3 * 3 + j / 3][num] = true;
}
}
}
helper(board, row, col, box, 0, 0);
}
private boolean helper(char[][]board, boolean[][]row, boolean[][]col, boolean[][]box, int i, int j){
if (j == 9) {
j = 0;
++i;
if (i == 9) return true;
}
if (board[i][j] == '.') {
for (int num = 0; num < 9; num++) {
boolean valid = !(row[i][num] || col[j][num] || box[i / 3 * 3 + j / 3][num]);
if (valid) {
row[i][num] = true;
col[j][num] = true;
box[i / 3 * 3 + j / 3][num] = true;
board[i][j] = (char)('1' + num);
if (helper(board, row, col, box, i, j + 1)) return true;
board[i][j] = '.';
row[i][num] = false;
col[j][num] = false;
box[i / 3 * 3 + j / 3][num] = false;
}
}
} else return helper(board, row, col, box, i, j + 1);
return false;
}
}
Python:
class Solution:
def solveSudoku(self, board: List[List[str]]) -> None:
"""
Do not return anything, modify board in-place instead.
"""
row, col, box = [[False for _ in range(9)] for _ in range(9)], [[False for _ in range(9)] for _ in range(9)], [[False for _ in range(9)] for _ in range(9)]
for i in range(9):
for j in range(9):
num = ord(board[i][j]) - ord('1')
if num > -1:
row[i][num], col[j][num], box[i // 3 * 3 + j // 3][num] = True, True, True
def trackback(board: List[List[str]], row: List[List[bool]], col: List[List[bool]], box: List[List[bool]], i: int, j: int) -> bool:
if j == 9:
j = 0
i += 1
if i == 9:
return True
if board[i][j] == '.':
for num in range(9):
valid = not (row[i][num] or col[j][num] or box[i // 3 * 3 + j // 3][num])
if valid:
row[i][num], col[j][num], box[i // 3 * 3 + j // 3][num], board[i][j] = True, True, True, chr(num + ord('1'))
if trackback(board, row, col, box, i, j + 1):
return True
board[i][j], row[i][num], col[j][num], box[i // 3 * 3 + j // 3][num] = '.', False, False, False
else:
return trackback(board, row, col, box, i, j + 1)
return False
trackback(board, row, col, box, 0, 0)
