LeetCode刷题--- 解数独

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力扣递归算法题

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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

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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

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解法

算法原理思路讲解 

为了存储每个位置的元素，我们需要定义⼀个⼆维数组。⾸先，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，同时检查⾏、列和九宫格是否唯⼀。

我们可以使⽤⼀个⼆维数组来记录每个数字在每⼀⾏中是否出现，⼀个⼆维数组来记录每个数字在每⼀列中是否出现。对于九宫格，我们可以以⾏和列除以 3 得到的商作为九宫格的坐标，并使⽤⼀个三维数组来记录每个数字在每⼀个九宫格中是否出现。在检查是否存在冲突时，只需检查⾏、列和九宫格⾥对应的数字是否已被标记。如果数字⾄少有⼀个位置（⾏、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，因此在找到⼀种可⾏的⽅法时，应该停⽌递

归，以防⽌正确的⽅法被覆盖。

（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>& board);

• 参数：board ；
• 返回值：布尔值 ；
• 函数作用：在当前坐标填⼊合适数字，查找数独答案。

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代码实现

class Solution {
public:
    bool checkRow[9][10];
    bool checkCol[9][10];
    bool gird[3][3][10];

    bool dfs(vector>& 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>& 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);
    }

};