0036_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.

JAVA

方法一

  并不知道如何来判断是不是一个可用的数独，只能尝试一下判断已经填写的数字中是否有不满足数独规则的，分别判断行、列、子格。没想到竟然过了，就是效率不高，在3/4左右。。

public class Solution {    static int MAXSIZE = 9;    static int SUBBOXSIZE = 3;    public boolean isValidSudoku(char[][] board) {        if (board.length != MAXSIZE || board[0].length != MAXSIZE){            return false;        }        for (int i = 0; i < MAXSIZE; ++i){            if (!judgeRow(board,i)){                return false;            }        }        for (int i = 0; i < MAXSIZE; ++i){            if (!judgeColumn(board,i)){                return false;            }        }        for (int i = 0; i < MAXSIZE; i += 3){            for(int j = 0; j < MAXSIZE; j += 3){                if (!judgeSubBox(board,i,j)){                    return false;                }            }        }        return true;    }    public boolean judgeRow(char[][] board,int row){        HashSet<Character> used = new HashSet<Character>();        for(int i = 0; i < MAXSIZE; ++i){            if(board[row][i] != '.' && used.contains(board[row][i])){                return false;            }else{                used.add(board[row][i]);            }        }        return true;    }    public boolean judgeColumn(char[][] board,int column){        HashSet<Character> used = new HashSet<Character>();        for(int i = 0; i < MAXSIZE; ++i){            if(board[i][column] != '.' && used.contains(board[i][column])){                return false;            }else{                used.add(board[i][column]);            }        }        return true;    }    public boolean judgeSubBox(char[][] board,int row,int column){        HashSet<Character> used = new HashSet<Character>();        for(int i = 0; i < SUBBOXSIZE; ++i){            for (int j = 0; j < SUBBOXSIZE; ++j){                if(board[row + i][column + j] != '.' && used.contains(board[row + i][column + j])){                    return false;                }else{                    used.add(board[row + i][column + j]);                }            }        }        return true;    }}

方法二