(回溯) LeetCode 37. 解数独

原题链接

一. 题目描述

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

二. 解题思路

本题是回溯中较难的一个题目，需要你通过回溯的方法实现解数独，玩过数独的同学一定十分清楚，数独就是同行同列不能有相同的数字出现，然后同九宫格不能有相同的数字出现。

这道题看似抽象，好像和上一道的N皇后千差万别，但是做起来思路都是很简单的，N皇后的意思是每行每列以及对角线只能由一个Q，我们使用一层循环来遍历每一行，然后通过一个isValid 函数判断当前位置放置皇后是否合法，当然本题也十分简单，只不过这个题需要你通过两层for 循环去遍历每一个位置，然后判断每一个位置放置某一数字是否合法，话不多说，我们进行回溯的三件套！

1. 确定递归参数：由于题目中只需要返回一个board 的二维字符数组，所以我们只需要传入一个board 数组就可以了，但是这个题目不像之前做的一样需要收集行多种结果，这个是只要有符合条件的就直接返回，所以我们将返回值类型定为bool 方便判断。

2. 递归终止条件：本题没有。

3. 单层循环：我们需要两层循环来遍历board 数组，外层遍历行，内层遍历列，然后判断每一个点是否为空白格，即（'.'） ，如果是，就再加一个for 循环，从1 到9 去枚举符合条件的答案，然后再去递归，回溯，就可以了，如果遍历了9 个数都没有找到符合条件的答案，那么直接return false 即可。

下图是代码随想录中的一个对递归以及回溯总结的流程图，大家可以了解一下。

话不多说！！！上代码！！

三. 代码

class Solution {
public:
    bool isValid(int row, int col, int val, vector<vector<char>>& board){
        // 判断同行
        for(int i = 0; i < 9; i++){
            if(board[row][i] == val) return false;
        }
        // 判断同列
        for(int i = 0; i < 9; i++){
            if(board[i][col] == val) return false;
        }
        // 判断同九宫格
        int startrow = (row / 3) * 3;
        int startcol = (col / 3) * 3;
        for(int i = startrow; i < startrow + 3; i++){
            for(int j = startcol; j < startcol + 3; j++){
                if(board[i][j] == val) return false;
            }
        }
        return true;
    }

    bool back(vector<vector<char>>& board){
        for(int i = 0; i < board.size(); i++){
            for(int j = 0; j < board[0].size(); j++){
                if(board[i][j] == '.'){
                    for(char k = '1'; k <= '9'; k++){       // 九个数字一个一个试
                        if(isValid(i, j, k, board)){
                            board[i][j] = k;
                            if(back(board)) return true;
                            board[i][j] = '.';
                        }
                    }
                    return false;       // 九个数都试完了还没答案就说明没结果
                }
            }
        }
        return true;
    }
    void solveSudoku(vector<vector<char>>& board) {
        back(board);
    }
};

四. 总结

本题是回溯算法中的难题了，大家可以先看我之前的题目，然后练习会之前的题目之后再看这题会容易很多。

时间复杂度：O(1)；

空间复杂度：O(1)。

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