前往小程序,Get更优阅读体验!
立即前往
首页
学习
活动
专区
工具
TVP
发布
社区首页 >专栏 >有效的数独

有效的数独

作者头像
狼啸风云
发布2023-10-23 16:13:09
1480
发布2023-10-23 16:13:09
举报
文章被收录于专栏:计算机视觉理论及其实现

请你判断一个 9 x 9 的数独是否有效。只需要 根据以下规则 ,验证已经填入的数字是否有效即可。

  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)

注意:

  • 一个有效的数独(部分已被填充)不一定是可解的。
  • 只需要根据以上规则,验证已经填入的数字是否有效即可。
  • 空白格用 '.' 表示。

示例 1:

代码语言:javascript
复制
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:true

示例 2:

代码语言:javascript
复制
输入:board = 
[["8","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:false
解释:除了第一行的第一个数字从 5 改为 8 以外,空格内其他数字均与 示例1 相同。 但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。

提示:

  • board.length == 9
  • board[i].length == 9
  • board[i][j] 是一位数字(1-9)或者 '.'

有效的数独满足以下三个条件:

    同一个数字在每一行只能出现一次;

    同一个数字在每一列只能出现一次;

    同一个数字在每一个小九宫格只能出现一次。

可以使用哈希表记录每一行、每一列和每一个小九宫格中,每个数字出现的次数。只需要遍历数独一次,在遍历的过程中更新哈希表中的计数,并判断是否满足有效的数独的条件即可。

对于数独的第

i
i

行第

j
j

列的单元格,其中

0 \le i, j < 90
0 \le i, j < 90

,该单元格所在的行下标和列下标分别为

i
i

j
j

,该单元格所在的小九宫格的行数和列数分别为

\Big\lfloor \dfrac{i}{3} \Big\rfloor
\Big\lfloor \dfrac{i}{3} \Big\rfloor

\Big\lfloor \dfrac{j}{3} \Big\rfloor
\Big\lfloor \dfrac{j}{3} \Big\rfloor

,其中

0\le \Big\lfloor \dfrac{i}{3} \Big\rfloor, \Big\lfloor \dfrac{j}{3} \Big\rfloor < 3
0\le \Big\lfloor \dfrac{i}{3} \Big\rfloor, \Big\lfloor \dfrac{j}{3} \Big\rfloor < 3

由于数独中的数字范围是

1
1

9
9

,因此可以使用数组代替哈希表进行计数。

具体做法是,创建二维数组

\textit{rows}
\textit{rows}

\textit{columns}
\textit{columns}

分别记录数独的每一行和每一列中的每个数字的出现次数,创建三维数组\textit{subboxes}记录数独的每一个小九宫格中的每个数字的出现次数,其中

\textit{rows}[i][\textit{index}]
\textit{rows}[i][\textit{index}]

\textit{columns}[j][\textit{index}]
\textit{columns}[j][\textit{index}]

[index]\textit{subboxes}\Big[\Big\lfloor \dfrac{i}{3} \Big\rfloor\Big]\Big[\Big\lfloor \dfrac{j}{3} \Big\rfloor\Big]\Big[\textit{index}\Big]
[index]\textit{subboxes}\Big[\Big\lfloor \dfrac{i}{3} \Big\rfloor\Big]\Big[\Big\lfloor \dfrac{j}{3} \Big\rfloor\Big]\Big[\textit{index}\Big]

分别表示数独的第

i
i

行第

j
j

列的单元格所在的行、列和小九宫格中,数字

\textit{index} + 1
\textit{index} + 1

出现的次数,其中

0\le \textit{index} < 90
0\le \textit{index} < 90

,对应的数字

\textit{index} + 1
\textit{index} + 1

满足

1 \le \textit{index} + 1 \le 9
1 \le \textit{index} + 1 \le 9

如果

\textit{board}[i][j]
\textit{board}[i][j]

填入了数字

n
n

,则将

\textit{rows}[i][n - 1]
\textit{rows}[i][n - 1]

\textit{columns}[j][n - 1]
\textit{columns}[j][n - 1]

\textit{subboxes}\Big[\Big\lfloor \dfrac{i}{3} \Big\rfloor\Big]\Big[\Big\lfloor \dfrac{j}{3} \Big\rfloor\Big]\Big[n - 1\Big]
\textit{subboxes}\Big[\Big\lfloor \dfrac{i}{3} \Big\rfloor\Big]\Big[\Big\lfloor \dfrac{j}{3} \Big\rfloor\Big]\Big[n - 1\Big]

各加

1
1

。如果更新后的计数大于

1
1

,则不符合有效的数独的条件,返回

\text{false}
\text{false}

如果遍历结束之后没有出现计数大于1的情况,则符合有效的数独的条件,返回

\text{true}
\text{true}

代码语言:javascript
复制
class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        int rows[9][9];
        int columns[9][9];
        int subboxes[3][3][9];
        
        memset(rows,0,sizeof(rows));
        memset(columns,0,sizeof(columns));
        memset(subboxes,0,sizeof(subboxes));
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                char c = board[i][j];
                if (c != '.') {
                    int index = c - '0' - 1;
                    rows[i][index]++;
                    columns[j][index]++;
                    subboxes[i / 3][j / 3][index]++;
                    if (rows[i][index] > 1 || columns[j][index] > 1 || subboxes[i / 3][j / 3][index] > 1) {
                        return false;
                    }
                }
            }
        }
        return true;
    }
};
本文参与 腾讯云自媒体同步曝光计划,分享自作者个人站点/博客。
原始发表:2023-10-19,如有侵权请联系 cloudcommunity@tencent.com 删除

本文分享自 作者个人站点/博客 前往查看

如有侵权,请联系 cloudcommunity@tencent.com 删除。

本文参与 腾讯云自媒体同步曝光计划  ,欢迎热爱写作的你一起参与!

评论
登录后参与评论
0 条评论
热度
最新
推荐阅读
领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档