题目
编写一个程序，通过填充空格来解决数独问题。
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。（请参考示例图）
数独部分空格内已填入了数字，空白格用 '.' 表示。
示例 1：
输入：board = [["5","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".",".","."],[".","9","8",".",".",".",".","6","."],["8",".",".",".","6",".",".",".","3"],["4",".",".","8",".","3",".",".","1"],["7",".",".",".","2",".",".",".","6"],[".","6",".",".",".",".","2","8","."],[".",".",".","4","1","9",".",".","5"],[".",".",".",".","8",".",".","7","9"]] 输出：[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]] 解释：输入的数独如上图所示，唯一有效的解决方案如下所示：
提示：文章来源：https://www.toymoban.com/news/detail-648827.html
board.length == 9board[i].length == 9-
board[i][j]是一位数字或者'.' - 题目数据 保证 输入数独仅有一个解
代码
可以参考前身：有效的数独文章来源地址https://www.toymoban.com/news/detail-648827.html
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>
void solveSudoku(char **board, int boardSize, int *boardColSize);
bool isValidSudoku(char **board, int boardSize, int *boardColSize);
int main()
{
// char b[9][10] =
// {"53..7....",
// "6..195...",
// ".98....6.",
// "8...6...3",
// "4..8.3..1",
// "7...2...6",
// ".6....28.",
// "...419..5",
// "....8..79"};
char b[9][10]=
{"..9748...",
"7........",
".2.1.9...",
"..7...24.",
".64.1.59.",
".98...3..",
"...8.3.2.",
"........6",
"...2759.."};
int t, *te;
char **board = (char **)malloc(sizeof(char *) * 9);
for (int i = 0; i < 9; i++)
{
board[i] = b[i];
printf("%s ", b[i]);
}
printf("\n");
solveSudoku(board, t, te);
for (int i = 0; i < 9; i++)
{
printf("%s ", board[i]);
}
return 0;
}
void solveSudoku(char **board, int boardSize, int *boardColSize)
{
int r, c;
for (int p = 0; p < 9; p++)
{
for (int q = 0; q < 9; q++)
{
if (board[p][q] == '.')
{
r = p;
c = q;
continue;
}
}
}
static int sign = 0;
for (int i = 0; i < 9; i++)
{
for (int j = 0; j < 9; j++)
{
if (board[i][j] == '.')
{
int p;
for (p = 1; p <= 9 && sign == 0; p++)
{
board[i][j] = p + '0';
if (isValidSudoku(board, boardSize, boardColSize) == 1)
{
if (i == r && j == c)
{
sign = 1;
}
solveSudoku(board, boardSize, boardColSize);
if (sign == 1)
{
return;
}
}
else
{
sign = 0;
continue;
}
}
if (p == 10)
{
board[i][j] = '.';
}
return;
}
}
}
for (int i = 0; i < 9; i++)
{
printf("%s ", board[i]);
}
printf("\n");
}
bool isValidSudoku(char **board, int boardSize, int *boardColSize)
{
int rownums[10], colnums[10];
memset(rownums, 0, sizeof(rownums));
memset(colnums, 0, sizeof(colnums));
for (int i = 0; i < 9; i++)
{
for (int j = 0; j < 9; j++)
{
if (board[i][j] != '.')
{
int number = board[i][j] - '0';
if (rownums[number] == 0)
{
rownums[number] = 1;
}
else
return false;
}
if (board[j][i] != '.')
{
int number = board[j][i] - '0';
if (colnums[number] == 0)
{
colnums[number] = 1;
}
else
return false;
}
}
memset(rownums, 0, sizeof(rownums));
memset(colnums, 0, sizeof(colnums));
}
int i = 0, j = 0;
for (int p = 3; p <= 9; p = p + 3)
{
for (int q = 3; q <= 9; q = q + 3)
{
i = p - 3;
for (; i < p; i++)
{
j = q - 3;
for (; j < q; j++)
{
if (board[i][j] != '.')
{
int number = board[i][j] - '0';
if (rownums[number] == 0)
{
rownums[number] = 1;
}
else
return false;
}
}
}
memset(rownums, 0, sizeof(rownums));
}
}
return true;
}
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