LeetCode Valid Sudoku

最新推荐文章于 2024-01-06 14:41:27 发布

kgduu

最新推荐文章于 2024-01-06 14:41:27 发布

阅读量370

点赞数

CC 4.0 BY-SA版权

分类专栏： OJ

本文链接：https://blog.youkuaiyun.com/xiexingshishu/article/details/50663507

OJ 专栏收录该内容

986 篇文章

订阅专栏

本文介绍了一种根据数独规则验证数独块合法性的算法，通过遍历每一行、每一列和每一个小矩形块，检查已填入的数字是否重复来判断数独是否合法。算法详细解释了其工作原理并提供了实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

A partially filled sudoku which is valid.

Note:

A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

题意：给出一个数独块，已经填了数据，问根据已填的数据判断数独是不是合法的。

思路：在扫描时，如果当前单元格已经填了数，就判断当前行、当前列、所在的小矩形块的已填的数是不是有重复。如果有重复，说明是不合法的。

代码如下：

class Solution
{
    public boolean isValidSudoku(char[][] board)
    {
        int n = board.length;
        Set<Integer> hs = new HashSet<Integer>();

        int m = (int)Math.floor(Math.sqrt(n));
        for (int i = 1; i <= n; i++)
        {
            hs.add(i);
        }

        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (board[i][j] != '.')
                {
                    Set<Integer> tmp = new HashSet<Integer>(hs);

                    for (int k = 0; k < n; k++)
                    {
                        if (board[i][k] == '.') continue;

                        int num = board[i][k] - '0';
                        if (!tmp.contains(num)) return false;
                        tmp.remove(num);
                    }

                    tmp = new HashSet<Integer>(hs);
                    for (int k = 0; k < n; k++)
                    {
                        if (board[k][j] == '.') continue;
                        int num = board[k][j] - '0';
                        if (!tmp.contains(num)) return false;
                        tmp.remove(num);
                    }

                    tmp = new HashSet<Integer>(hs);
                    int row = i / m, col = j / m;
                    for (int a = row * m; a < (row + 1) * m; a++)
                    {
                        for (int b = col * m; b < (col + 1) * m; b++)
                        {
                            if (board[a][b] == '.') continue;
                            int num = board[a][b] - '0';
                            if (!tmp.contains(num)) return false;
                            tmp.remove(num);
                        }
                    }
                    
                }
            }
        }
        return true;
    }
}