[LeetCode] 128: Valid Sudoku

[Problem]

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.

[Analysis]
最开始以为需要判断是不是一个可解的数独，然后TLE。。。原来可以不用判断是否可解，只需要判断当前数独状态是否合法即可。

[Solution]

class Solution {
public:
bool isValidSudoku(vector<vector<char> > &board) {
// Note: The Solution object is instantiated only once and is reused by each test case.

// initial
bool rowUsed[9][10], columnUsed[9][10], cubeUsed[9][10];
memset(rowUsed, 0, sizeof(rowUsed));
memset(columnUsed, 0, sizeof(columnUsed));
memset(cubeUsed, 0, sizeof(cubeUsed));

// set rowUsed, columnUsed, cubeUsed
vector<pair<int, int> > unSolved;
for(int i = 0; i < 9; ++i){
for(int j = 0; j < 9; ++j){
// filled
if(board[i][j] != '.'){
// set rowUsed
if(rowUsed[i][board[i][j]-'0'])return false;
rowUsed[i][board[i][j]-'0'] = true;

// set columnUsed
if(columnUsed[j][board[i][j]-'0'])return false;
columnUsed[j][board[i][j]-'0'] = true;

// set cubeUsed
if(cubeUsed[i/3*3+j/3][board[i][j]-'0'])return false;
cubeUsed[i/3*3+j/3][board[i][j]-'0'] = true;
}
// empty
else{
unSolved.push_back(pair<int, int>(i, j));
}
}
}

return true;
}
};

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