Leetcode# 36 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

'.'

.



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.

Difficulty:Easy

遍历就好了，没什么难的。

bool isValidSudoku(vector<vector<char> >& board) {
int row[9],line[9],sq[9];
int temp;
for(int i = 0;i<9;i++)
{
memset(row,0,sizeof(row));
for(int j = 0;j<9;j++)
{
if(board[i][j]=='.')
continue;
temp = board[i][j]-'0';
row[temp-1]++;
}
for(int j = 0;j<9;j++)
{
if(row[j]>1)
return false;
}
}

for(int i = 0;i<9;i++)
{
memset(line,0,sizeof(line));
for(int j = 0;j<9;j++)
{
if(board[j][i]=='.')
continue;
temp = board[j][i]-'0';
line[temp-1]++;
}
for(int j = 0;j<9;j++)
{
if(line[j]>1)
return false;
}
}

int a = 0, b = 0,cou=0;
while(cou<9)
{memset(sq,0,sizeof(sq));
switch(cou){
case 0:
a=0;b=0;
break;
case 1:
a=0;b=3;
break;
case 2:
a=0;b=6;
break;
case 3:
a=3;b=0;
break;
case 4:
a=3;b=3;
break;
case 5:
a=3;b=6;
break;

case 6:
a=6;b=0;
break;
case 7:
a=6;b=3;
break;
case 8:
a=6;b=6;
break;
}
for(int i=a;i<a+3;i++)
{
for(int j=b;j<b+3;j++)
{
if(board[i][j]=='.')
continue;
temp = board[i][j]-'0';
sq[temp-1]++;
}
}
for(int j = 0;j<9;j++)
{
if(sq[j]>1)
return false;
}
cou++;
}

return true;
}