☆打卡算法☆LeetCode 37、解数独 算法解析

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一、题目

1、算法题目

“编写程序，填写数独剩余空格，解数独。”

题目链接：

来源：力扣（LeetCode）

链接：37. 解数独 - 力扣（LeetCode） (leetcode-cn.com)

2、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。

示例图：

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：

二、解题

1、思路分析

这个题考虑按行解题的顺序，通过递归+回溯枚举所有可能的填法，当递归到最后的时候，仍然是合理的答案，说明我们找到了答案。

在递归的过程中，发现当前格子不能填下任何一个数字，那么就进行回溯。

2、代码实现

代码参考：

public class Solution 
{
    //九列
    List<Dictionary<int, bool>> liCol = new List<Dictionary<int, bool>>();
    //九行
    List<Dictionary<int, bool>> liRow = new List<Dictionary<int, bool>>();
    //3*3格子 九个
    List<Dictionary<int, bool>> liBox = new List<Dictionary<int, bool>>();
    char[][] boardAllres = null;
    public void SolveSudoku(char[][] board) 
    {
        //判断一共有多少个数字
        int hasNumCount = 0;
        Dictionary<int, bool> dicAllTemp = new Dictionary<int, bool>();
        //创建一个键值对集合存储数字是否出现过 
        for (int i = 1; i <= 9; i++)
        {
            dicAllTemp.Add(i, false);
        }
        for (int i = 0; i < 9; i++)
        {
            //存储对应的九列
            liCol.Add(new Dictionary<int, bool>(dicAllTemp));
            //存储对应的九行
            liRow.Add(new Dictionary<int, bool>(dicAllTemp));
            //存储对应的九个3*3格子
            liBox.Add(new Dictionary<int, bool>(dicAllTemp));
        }
        //遍历整个集合 将已出现的数字赋值为true 
        for (int i = 0; i < board.Length; i++)
        {
            for (int j = 0; j < board[0].Length; j++)
            {
                if (board[i][j] != '.')
                {
                    int temp =  board[i][j]-48;
                    liRow[i][temp] = true;
                    liCol[j][temp] = true;
                    int count = (i / 3)*3 + j / 3;
                    liBox[count][temp] = true;
                    hasNumCount++;
                }
            }
        }
        Recursive(board, hasNumCount, 0, 0);
        board= boardAllres;
    }

    public bool Recursive(char[][] board, int hasNumCount,int i,int j)
    {
        //81个数字代表已经填满了 直接返回结果
        if (hasNumCount==81)
        {
            boardAllres = board;
            return true;
        }
        for (; i < board.Length; i++)
        {
            for (j=0; j < board[0].Length; j++)
            {
                if (board[i][j] == '.')
                {
                    //查出该行对应的所有未使用的数字 一次遍历填入
                    List<int> liRes = liRow[i].Where(e => e.Value == false).Select(e => e.Key).ToList();
                    foreach (var item in liRes)
                    {
                        //判读位于那个格子 
                        int count = (i / 3) * 3 + j / 3;
                        if (liCol[j][item] == false && liBox[count][item] == false && liRow[i][item] == false)
                        {
                            //总数+1 对应的位置赋值为true
                            hasNumCount++;
                            liRow[i][item] = true;
                            liCol[j][item] = true;
                            liBox[count][item] = true;
                            //之所以加48为了将数字转为char类型存进去
                            board[i][j] =(char)(item+48);
                            //递归 存在不符合就返回 结果为true只有一种情况 那就是已经完成了
                            if (Recursive(board, hasNumCount, i, 0))
                            {
                                return true;
                            }
                            //不符合 将操作撤回到之前 
                            hasNumCount--;
                            board[i][j] = '.';
                            liRow[i][item] = false;
                            liCol[j][item] = false;
                            liBox[count][item] = false;
                        }
                    }
                    return false;
                }
            }
        }
        return false;
    }
}

3、时间复杂度

时间复杂度 ：

空间复杂度：

三、总结

遍历过程之后，进行递归和回溯枚举，这个还是很难的。

要思考好边界问题，还有回溯到当前递归层时，其他变量的复原问题。