### 问题描述

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.

### 代码

class Solution {
public:
bool isValidSudoku(vector<vector<char> >& board) {
int m = board.size();
int n = board[0].size();
for(int i=0;i<9;i++) {
bool exists1[10]={
0
};
bool exists2[10]={
0
};
for(int j=0;j<9;j++) {
// 横的
char v1 = board[i][j];
if(v1!='.'&&exists1[v1-'0']) {
return false;
} else if(v1!='.') {
exists1[v1-'0']=true;
}
// 纵的
char v2 = board[j][i];
if(v2!='.'&&exists2[v2-'0']) {
return false;
} else if(v2!='.') {
exists2[v2-'0']=true;
}
}
}
for(int i=0;i<3;i++) {
for(int j=0;j<3;j++){
// 遍历九宫格
int x=i*3;
int y=j*3;
bool exists[10]={
0
};
for(int p=0;p<3;p++) {
for(int q=0;q<3;q++) {
char v = board[x+p][y+q];
if(v!='.'&&exists[v-'0']) {
return false;
} else if(v!='.') {
exists[v-'0']=true;
}
}
}
}
}
return true;
}
};


int main() {
vector<vector<char> > board(9);
string strs[9] = {".87654321","2........","3........","4........","5........","6........","7........","8........","9........"};
for(int i=0;i<9;i++) {
vector<char> vec(9);
for(int j=0;j<9;j++) {
char ch = strs[i][j];
vec[j]=ch;
}
board[i]=vec;
}
Solution s;
cout<<s.isValidSudoku(board)<<endl;
}