[Leetcode] valid sudoku 有效数独
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.
题意:判断矩阵是否是一个数独矩阵。所谓的数独矩阵就是9*9的矩阵,每一行一个元素只出现一次、每一列一个元素只出现一次,并且在任意一个小矩阵(3×3)中一个元素只出现一次(所说的元素指0-9这九个数字)。a)第i个九宫格的第j个格点的行号可表示为i/3*3+j/3;b)第i个九宫格的第j个格点的列号可表示为i%3*3+j%3(详见陆草纯的博客)。整体的思路是:开辟三个二维标志数组,m[i][j]表示当前的这个位置,是否被访问过。
代码一陆草纯的博客,代码如下:
class Solution {
public:
bool isValidSudoku(vector<vector<char> > &board) {
for(int i = ; i < ; i ++)
{
unordered_map<char, bool> m1; //check i_th row
unordered_map<char, bool> m2; //check i_th column
unordered_map<char, bool> m3; //check i_th subgrid
for(int j = ; j < ; j ++)
{
if(board[i][j] != '.')
{
if(m1[board[i][j]] == true)
return false;
m1[board[i][j]] = true;
}
if(board[j][i] != '.')
{
if(m2[board[j][i]] == true)
return false;
m2[board[j][i]] = true;
}
if(board[i/*+j/][i%*+j%] != '.')
{
if(m3[board[i/*+j/][i%*+j%]] == true)
return false;
m3[board[i/*+j/][i%*+j%]] = true;
}
}
}
return true;
}
};
因为只要考虑出现的数字是否有重复的,所以,只需用当前元素的值转化为坐标信息,访问就行。Grandyang代码二:
class Solution {
public:
bool isValidSudoku(vector<vector<char> > &board)
{
int row=board.size(),col=board[].size();
if(row !=||col !=) return false; vector<vector<bool>> rowFlag(row,vector<bool>(col,false));
vector<vector<bool>> colFlag(row,vector<bool>(col,false));
vector<vector<bool>> cellFlag(row,vector<bool>(col,false)); for(int i=;i<row;++i)
{
for(int j=;j<col;++j)
{
if(board[i][j]>=''&&board[i][j]<='')
{
int c=board[i][j]-'';
if(rowFlag[i][c]||colFlag[c][j]||cellFlag[*(i/)+j/][c])
return false; rowFlag[i][c]=true;
colFlag[c][j]=true;
cellFlag[*(i/)+j/][c]=true;
}
}
}
return true;
}
};
想把代码一按照代码二那样写,总是通不过,抽时间再看看问题出在哪。
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