leetCode 36.Valid Sudoku(有效的数独) 解题思路和方法

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.

思路：题目很简单，主要是规则的理解，数独的游戏没有玩过，不知道什么规则，我以为任意9个方格1-9的个数都至多为1，谁知规则是特定的九个格内1-9的个数至多为1，其他不考虑。代码比较啰嗦，但思路清晰，如下：

public class Solution {
    //置为静态变量
    static Map map = new HashMap();
    public boolean isValidSudoku(char[][] board) {
        //判断每行
        for(int i = 0; i < board.length; i++){
            initMap();//每次均需初始化
            for(int j = 0; j < board[0].length; j++){
                //是数字
                if(board[i][j] >= '0' && board[i][j] <= '9'){
                    if(map.get(board[i][j]) > 0){//说明重复数字
                        return false;
                    }else{
                        map.put(board[i][j],1);
                    }
                }else if(board[i][j] != '.'){//出现空格和0-9之外的字符
                    return false;//直接返回false
                }
            }
        }
        //判断每列
        for(int i = 0; i < board[0].length; i++){
            initMap();//每次均需初始化
            for(int j = 0; j < board.length; j++){
                //是数字
                if(board[j][i] >= '0' && board[j][i] <= '9'){
                    if(map.get(board[j][i]) > 0){//说明重复数字
                        return false;
                    }else{
                        map.put(board[j][i],1);
                    }
                }else if(board[j][i] != '.'){//出现空格和0-9之外的字符
                    return false;//直接返回false
                }
            }
        }
        //判断九宫格
        for(int i = 0; i < board.length - 2; i = i+3){//行{
            for(int j = 0; j < board[0].length - 2; j=j+3){
                initMap();//初始化
                for(int m = i; m < i + 3;m++){
                    for(int n = j; n < j+3; n++){
                        //是数字
                        if(board[m][n] >= '0' && board[m][n] <= '9'){
                            if(map.get(board[m][n]) > 0){//说明重复数字
                                return false;
                            }else{
                                map.put(board[m][n],1);
                            }
                        }else if(board[m][n] != '.'){//出现空格和0-9之外的字符
                            return false;//直接返回false
                        }
                    }
                }
            }
        }
        return true;
    }
    //初始化map为每个key均赋值0
    private void initMap(){
        for(char i = '0';i <= '9'; i++){
            map.put(i,0);
        }
    }
}