“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)
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:
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:
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。
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;
}
}
时间复杂度 :
空间复杂度:
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。