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  • 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:题目非常easy。主要是规则的理解,数独的游戏没有玩过。不知道什么规则,我以为随意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。其它不考虑。

    代码比較啰嗦,但思路清晰,例如以下:

    public class Solution {
        //置为静态变量
        static Map<Character,Integer> map = new HashMap<Character,Integer>();
        public boolean isValidSudoku(char[][] board) {
            //推断每行
            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){//说明反复数字
                            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
                    }
                }
            }
            //推断每列
            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){//说明反复数字
                            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
                    }
                }
            }
            //推断九宫格
            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){//说明反复数字
                                    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为每一个key均赋值0
        private void initMap(){
            for(char i = '0';i <= '9'; i++){
                map.put(i,0);
            }
        }
    }


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  • 原文地址:https://www.cnblogs.com/cxchanpin/p/7044200.html
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