[LeetCode]Valid Sudoku

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

'.'

.

LeetCode Source

分析：利用map来判断

Brute Force解法。

1）每一行是否有重复的数；

2）每一列是否有重复的数；

3）每一个9宫是否有重复的数。

可以利用Map来判断是否重复。

class Solution {
public:
bool isValidSudoku(vector<vector<char> > &board) {
vector<map<char,int>> Sudoku(27);
for(int i=0;i<9;++i){ //行相等判断
for(int j=0;j<9;++j){
if(board[i][j]=='.')
continue;
if(board[i][j]>'9'||board[i][j]<'0')
return false;
if((Sudoku[i])[board[i][j]]>0)
return false;
(Sudoku[i])[board[i][j]]++;
}
}

for(int i=0;i<9;++i){//列相等判断
for(int j=0;j<9;++j){
if(board[j][i]=='.')
continue;
if((Sudoku[i+9])[board[j][i]]>0)
return false;
(Sudoku[i+9])[board[j][i]]++;
}
}

int it=0; //每个单元格相等判断
while(it<9){//迭代9次
for(int i=(it%3)*3;i<(it%3)*3+3;++i){
for(int j=(it/3)*3;j<(it/3)*3+3;++j){
if(board[i][j]=='.')
continue;
if((Sudoku[it+18])[board[i][j]]>0)
return false;
(Sudoku[it+18])[board[i][j]]++;
}
}
++it;
}
return true;
}
};