LeetCode刷题--- 解数独
2023-12-29 09:29:08
个人主页:元清加油_【C++】,【C语言】,【数据结构与算法】-CSDN博客
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力扣递归算法题
【C++】? ??
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数据结构与算法
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前言:这个专栏主要讲述递归递归、搜索与回溯剪枝算法,所以下面题目主要也是这些算法做的 ?
我讲述题目会把讲解部分分为3个部分:
1、题目解析
2、算法原理思路讲解
3、代码实现
解数独
题目链接:解数独
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:
board.length == 9
board[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
解法
算法原理思路讲解?
为了存储每个位置的元素,我们需要定义?个?维数组。?先,我们记录所有已知的数据,然后遍历所有需要处理的位置,并遍历数字 1~9。对于每个位置,我们检查该数字是否可以存放在该位置,同时检查?、列和九宫格是否唯?。
我们可以使??个?维数组来记录每个数字在每??中是否出现,?个?维数组来记录每个数字在每?列中是否出现。对于九宫格,我们可以以?和列除以 3 得到的商作为九宫格的坐标,并使??个三维数组来记录每个数字在每?个九宫格中是否出现。在检查是否存在冲突时,只需检查?、列和九宫格?对应的数字是否已被标记。如果数字?少有?个位置(?、列、九宫格)被标记,则存在冲突,因此不能在该位置放置当前数字。
特别地,在本题中,我们需要直接修改给出的数组,因此在找到?种可?的?法时,应该停?递
归,以防?正确的?法被覆盖。
(1)全局变量
bool checkRow[9][10];
bool checkCol[9][10];
bool gird[3][3][10];
- checkRow(用于判断行是否有重复)
- checkCol(用于判断列是否有重复)
- gird(用于判断 3x3 是否有重复)
(2)设计递归函数
bool dfs(vector<vector<char>>& board);
- 参数:board ;
- 返回值:布尔值 ;
- 函数作用:在当前坐标填?合适数字,查找数独答案。
代码实现
class Solution {
public:
bool checkRow[9][10];
bool checkCol[9][10];
bool gird[3][3][10];
bool dfs(vector<vector<char>>& board)
{
for (int i = 0; i < 9; i++)
{
for (int j = 0; j < 9; j++)
{
if (board[i][j] == '.')
{
for (int m = 1; m <= 9; m++)
{
if (!checkRow[i][m] && !checkCol[j][m] && !gird[i / 3][j / 3][m])
{
board[i][j] = m + '0';
checkRow[i][m] = checkCol[j][m] = gird[i / 3][j / 3][m] = true;
if (dfs(board) == true)
return true;
board[i][j] = '.';
checkRow[i][m] = checkCol[j][m] = gird[i / 3][j / 3][m] = false;
}
}
return false; // 到这说明上一步已经错了
}
}
}
return true;
}
void solveSudoku(vector<vector<char>>& board)
{
for (int i = 0; i < 9; i++)
{
for (int j = 0; j < 9; j++)
{
if (board[i][j] != '.')
{
int number = board[i][j] - '0';
checkRow[i][number] = checkCol[j][number] = gird[i / 3][j / 3][number] = true;
}
}
}
dfs(board);
}
};
文章来源:https://blog.csdn.net/weixin_74268082/article/details/135282514
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