一、题目描述

判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。

数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。

Leetcode No.36 有效的数独_i++
上图是一个部分填充的有效的数独。

数独部分空格内已填入了数字，空白格用 '.' 表示。

示例 1:
输入:
[
  ["5","3",".",".","7",".",".",".","."],
  ["6",".",".","1","9","5",".",".","."],
  [".","9","8",".",".",".",".","6","."],
  ["8",".",".",".","6",".",".",".","3"],
  ["4",".",".","8",".","3",".",".","1"],
  ["7",".",".",".","2",".",".",".","6"],
  [".","6",".",".",".",".","2","8","."],
  [".",".",".","4","1","9",".",".","5"],
  [".",".",".",".","8",".",".","7","9"]
]
输出: true


示例 2:
输入:
[
  ["8","3",".",".","7",".",".",".","."],
  ["6",".",".","1","9","5",".",".","."],
  [".","9","8",".",".",".",".","6","."],
  ["8",".",".",".","6",".",".",".","3"],
  ["4",".",".","8",".","3",".",".","1"],
  ["7",".",".",".","2",".",".",".","6"],
  [".","6",".",".",".",".","2","8","."],
  [".",".",".","4","1","9",".",".","5"],
  [".",".",".",".","8",".",".","7","9"]
]
输出: false

解释: 除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。
     但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。
说明:

一个有效的数独（部分已被填充）不一定是可解的。
只需要根据以上规则，验证已经填入的数字是否有效即可。
给定数独序列只包含数字 1-9 和字符 '.' 。
给定数独永远是 9x9 形式的。

二、解题思路

1、验证数字 1-9 在每一行只能出现一次。

2、验证数字 1-9 在每一列只能出现一次。

3、验证数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。

三、代码

package is_valid_sudoku;

import java.util.HashMap;
import java.util.Map;
public class Solution {
    public boolean isValidSudoku(char[][] board) {
        int len=board.length;
        for(int i=0;i<len;i++){
            if(!checkRow(board,i)){
                System.out.println("第"+i+"行验证失败");
                return false;
            }else{
                System.out.println("第"+i+"行验证成功");
            }
        }
        for(int i=0;i<len;i++){
            if(!checkColumn(board,i)){
                System.out.println("第"+i+"列验证失败");
                return false;
            }
            else{
                System.out.println("第"+i+"列验证成功");
            }
        }
        for(int i=0;i<3;i++) {
            for(int j=0;j<3;j++){
                if (!checkMatrix(board, i*3, j*3)) {
                    return false;
                }
                else{
                    System.out.println("第"+i*3+","+j*3+"矩阵验证成功");
                }
            }
        }
        return true;
    }
    public boolean checkRow(char[][] board,int row){
        int len=board.length;
        Map<Character, Integer> map=new HashMap<>();
        for(int i=0;i<10;i++){
            map.put(Character.forDigit(i,10),0);
        }
        for(int i=0;i<len;i++){
            if(board[row][i]!='.'){
                if(map.get(board[row][i])>=1){
                    return false;
                }else{
                    map.put(board[row][i],map.get(board[row][i])+1);
                }
            }
        }
        return true;
    }
    public boolean checkColumn(char[][] board,int column){
        int len=board.length;
        Map<Character, Integer> map=new HashMap<>();
        for(int i=0;i<10;i++){
            map.put(Character.forDigit(i,10),0);
        }
        for(int i=0;i<len;i++){
            if(board[i][column]!='.'){
                if(map.get(board[i][column])>=1){
                    return false;
                }else{
                    map.put(board[i][column],map.get(board[i][column])+1);
                }
            }
        }
        return true;
    }
    public boolean checkMatrix(char[][] board,int rowStart,int columnStart){
        int len=board.length;
        Map<Character, Integer> map=new HashMap<>();
        for(int n=0;n<10;n++){
            map.put(Character.forDigit(n,10),0);
        }
        int rowEnd=rowStart+3;
        int columnEnd=columnStart+3;
        for(int i=rowStart;i<rowEnd;i++){
            for(int j=columnStart;j<columnEnd;j++) {
                if(board[i][j]!='.'){
                    if (map.get(board[i][j]) >= 1) {
                        return false;
                    } else {
                        map.put(board[i][j], map.get(board[i][j]) + 1);
                        System.out.println(map);
                    }
                }
            }
        }
        System.out.println(map);
        return true;
    }

    public static void main(String[] args) {
        Solution solution=new Solution();
        char[][] board={
                {'5','3','.','.','7','.','.','.','.'},
                {'6','.','.','1','9','5','.','.','.'},
                {'.','9','8','.','.','.','.','6','.'},
                {'8','.','.','.','6','.','.','.','3'},
                {'4','.','.','8','.','3','.','.','1'},
                {'7','.','.','.','2','.','.','.','6'},
                {'.','6','.','.','.','.','2','8','.'},
                {'.','.','.','4','1','9','.','.','5'},
                {'.','.','.','.','8','.','.','7','9'}
        };
        System.out.println(solution.isValidSudoku(board));
    }

}

四、复杂度分析

时间复杂度：O(1)，只对 81 个单元格进行了3次迭代。

空间复杂度：O(1)，只需要常数空间存放若干变量。

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Leetcode No.36 有效的数独_i++_02