Valid Sudoku

难度：1

个人总结：傻傻地写错两个符号，1Y失败

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.

由于输入一定是9*9的数组，所以也没有什么要注意的地方

class Solution {public:bool row(vector<char> &r){int opt=0;for(int i=0;i<r.size();i++){if(r[i] == '.')continue;if(opt&(1<<(r[i]-'0'-1)))return false;opt+=(1<<(r[i]-'0'-1));}return true;}bool col_check(vector<vector<char> > &board, int col){int opt=0;for(int i=0;i<board.size();i++){if(board[i][col] == '.')continue;if(opt&(1<<(board[i][col]-'0'-1)))return false;opt+=(1<<(board[i][col]-'0'-1));}return true;}bool center_check(vector<vector<char> > &board, int x, int y){int dir[8][2]={{1,0},{1,-1},{1,1},{-1,-1},{-1,1},{-1,0},{0,1},{0,-1}};int opt=0;opt+=(1<<(board[x][y]-'0'-1));for(int i=0;i<8;i++){if(board[x+dir[i][0]][y+dir[i][1]] == '.')continue;if(opt&(1<<(board[x+dir[i][0]][y+dir[i][1]]-'0'-1)))return false;opt+=(1<<(board[x+dir[i][0]][y+dir[i][1]]-'0'-1));}return true;}    bool isValidSudoku(vector<vector<char> > &board) {for(int i=0;i<board.size();i++){if(!row(board[i]))return false;}if(board.size()>0&&board[0].size()>0){for(int i=0;i<board[0].size();i++){if(!col_check(board,i))return false;}}int center[9][2]={{1, 1}, {1, 4}, {1, 7}, {4, 1}, {4, 4}, {4, 7}, {7, 1}, {7, 4}, {7, 7} };for(int i=0;i<9;i++){if(!center_check(board,center[i][0],center[i][1]))return false;}return true;    }};