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.
Analysis:
first scan every row and column. then scan every square.
Java
public boolean isValidSudoku(char[][] board) { ArrayList<Character> row = new ArrayList<>(); ArrayList<Character> col = new ArrayList<>(); for(int i=0;i<9;i++){ row.clear(); col.clear(); for(int j=0;j<9;j++){ if(board[i][j]!='.'){ if(col.contains(board[i][j])) return false; else col.add(board[i][j]); } if(board[j][i]!='.'){ if(row.contains(board[j][i])) return false; else row.add(board[j][i]); } } } ArrayList<Character> block = new ArrayList<>(); for(int i=0;i<9;i=i+3){ for(int j=0;j<9;j=j+3){ block.clear(); for(int m=0;m<3;m++){ for(int n = 0;n<3;n++){ if(board[i+m][j+n]!='.') if(block.contains(board[i+m][j+n])) return false; else block.add(board[i+m][j+n]); } } } } return true; }c++
bool isValidSudoku(vector<vector<char> > &board) { for(int i=0; i<board.size(); i++){ for(int j=0; j<board[i].size();j++){ for(int k=j+1; k<board[i].size();k++){ if(board[i][j] == board[i][k] && board[i][j]!='.') return false; } for(int k=0; k<board.size(); k++){ if(board[i][j] == board[k][j] && board[i][j]!='.' && i!=k) return false; } } } int row[9]; for(int i=0; i<9;i+=3){ for(int j=0; j<9; j+=3){ memset(row,0,9*sizeof(int)); for(int m=0; m<3;m++){ for(int n=0; n<3;n++){ if(board[i+m][j+n] == '.') continue; if(row[board[m+i][j+n]-49] == 1) return false; row[board[m+i][j+n]-49]++; } } } } return true; }