LeetCode刷题--- 解数独

发布时间:2023年12月29日

个人主页:元清加油_【C++】,【C语言】,【数据结构与算法】-CSDN博客

个人专栏

力扣递归算法题

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【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

  1. 数字?1-9?在每一行只能出现一次。
  2. 数字?1-9?在每一列只能出现一次。
  3. 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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