[leetcode-36-Valid Sudoku]

Posted 2020-09-15 hellowOOOrld

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.

思路：

分别用三个二维数组分别记录行、列和子数独里面数字的使用情况。一旦发现已经有数字之前被占用了，返回false。

其中需要注意的是，子数独的下标，如上图中一共存在3*3个子数独，每行3个，一共3行。对于大的数独索引(i,j)，对应到

子数独的下标为：i/3 * 3 + j/3 。（比如，对于一个二维m*n的数组，坐标(a,b)映射到一维坐标为：所在行a乘以列数n+列坐标b 即a*n+b）

这里i/3 * 3 + j/3相当于第i/3行，j/3列。

如下是参考大神写的：

参考：

bool isValidSudoku(vector<vector<char>>& board)
 {
     int rowcheck[9][9] = {0},colcheck[9][9] = {0},subboxcheck[9][9] = {0};
     for(int i = 0;i<board.size();i++)
     {
         for(int j =0 ;j<board[0].size();j++)
         {
             if(board[i][j] != ‘.‘)
             {
                 int num = board[i][j] - ‘0‘ - 1,k = i/3 * 3 + j/3;//记录子数独的下标一共3*3个子数独 m*n的数组 (a,b) 一维下标为a*n+b
                 if(rowcheck[i][num] || colcheck[j][num] || subboxcheck[k][num])return false;
                 rowcheck[i][num] = colcheck[j][num]  = subboxcheck[k][num] = 1;
             }
         }
     }
     return true;
}

 

https://discuss.leetcode.com/topic/8241/my-short-solution-by-c-o-n2/2