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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.
Hash Table
这道题求的是就宫格是否是有效的，只是对于已经填好的数字，而不是有没有解的问题，这里需要考虑的是每一行每一列，还有每一个九宫格是否在已经填的数字中出现
了重复，所以想到的是两重循环，依次的往下遍历，而在这个过程中又三个判断，而采用三个数组来记录，行，列，九宫格，而这里九宫格是采用将i，j转换成第几个九宫格
#include<iostream>
#include<vector>
using namespace std;
int flag1[10];
int flag2[10][10];
int flag3[10][10];
int changeij(int i,int j)
{
int result;
if(i<=2)
result=j/3;
else if(i>2&&i<=5)
result=3+j/3;
else
result=6+j/3;
return result;
}
bool isValidSudoku(vector<vector<char>>& board) {
memset(flag2,0,sizeof(flag2));
memset(flag3,0,sizeof(flag3));
for(int i=0;i!=9;i++)
{
memset(flag1,0,sizeof(flag1));
for(int j=0;j!=9;j++)
{
if(board[i][j]!=‘.‘)
{
if(flag1[board[i][j]-48]==1)
return false;
else
flag1[board[i][j]-48]=1;
if(flag2[board[i][j]-48][j]==1)
return false;
else
flag2[board[i][j]-48][j]=1;
if(flag3[changeij(i,j)][board[i][j]-48]==1)
return false;
else
flag3[changeij(i,j)][board[i][j]-48]=1;
}
}
}
return true;
}
int main()
{
vector<vector<char> > vec=["....5..1.",".4.3.....",".....3..1","8......2.","..2.7....",".15......",".....2...",".2.9.....","..4......"];
cout<<changeij(4,2)<<endl;
}
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原文地址：http://www.cnblogs.com/yanliang12138/p/4470542.html
