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.
class Solution { public: bool isValidSudoku(vector<vector<char> > &board) { vector<bool> flag(10, false); //indicate the appearance of 0,1,2,..., 9 //check line for(int i = 0; i<9; i++) { for(int j = 0; j<9; j++) { if(board[i][j]==‘.‘) continue; if(flag[board[i][j]-‘0‘]) return false; flag[board[i][j]-‘0‘] = true; } flag.assign(board.size()+1,false); //reset the vector } flag.assign(board.size()+1,false); //check column for(int j = 0; j<9; j++) { for(int i = 0; i<9; i++) { if(board[i][j]==‘.‘) continue; if(flag[board[i][j]-‘0‘]) return false; flag[board[i][j]-‘0‘] = true; } flag.assign(board.size()+1,false); } flag.assign(board.size()+1,false); //check small square for(int i = 0; i < 9; i+=3){ for(int j = 0; j <9; j+=3){ for(int m = 0; m < 3; m++){ for(int n = 0; n < 3; n++){ if(board[i+m][j+n]==‘.‘) continue; if(flag[board[i+m][j+n]-‘0‘]) return false; flag[board[i+m][j+n]-‘0‘] = true; } } flag.assign(board.size()+1,false); } } return true; } };
如果没有‘.‘，那么我们可以用以下的方法判断是否valid （参数是某一行，某一列，或是某一个small square的9个元素）
bool check(vector<int> v) { sort(v.begin(), v.end()); for (int i = 0; i < (int)v.size(); ++i) { if (v[i] != i + 1) { return false; } } return true; }
原文：http://www.cnblogs.com/qionglouyuyu/p/4857030.html