Leetcode刷题100天—36. 有效的数独(数组+哈希)—day26

r囧r小猫 2022-09-12 02:54 189阅读 0赞

前言：

> 作者：神的孩子在歌唱
> 
> 大家好，我叫运智

![image-20210902204936105][]

#### [36. 有效的数独][36.] ####

难度中等575

请你判断一个 `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。

1.  数字 `1-9` 在每一行只能出现一次。
2.  数字 `1-9` 在每一列只能出现一次。
3.  数字 `1-9` 在每一个以粗实线分隔的 `3x3` 宫内只能出现一次。（请参考示例图）

数独部分空格内已填入了数字，空白格用 `'.'` 表示。

**注意：**

*  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。

**示例 1：**

![img][]

输入：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"]]
    输出：true

**示例 2：**

输入：board = 
    [["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。

**提示：**

*  `board.length == 9`
 *  `board[i].length == 9`
 *  `board[i][j]` 是一位数字或者 `'.'`

package 数组;
    
    import java.util.HashMap;
    import java.util.HashSet;
    import java.util.Set;
    
    import org.omg.CORBA.PUBLIC_MEMBER;
    
    public class _36_有效的数独 { 
    //通过哈希表判断是否有重复的数字，在通过for循环一个个遍历
        public boolean isValidSudoku(char[][] board) { 
    // 通过哈希映射,如果有数大于2的就返回false
    // 我们这里设置存储列的哈希，行的哈希和子独数的哈希
        	HashMap<Integer, Integer>[] rows=new HashMap[9];//行
        	HashMap<Integer, Integer>[] colums=new HashMap[9];//列
        	HashMap<Integer, Integer>[] boxes=new HashMap[9];//子独数
    // 由于这里是九行九列,所以我们要遍历一下给每个都设置一个哈希空间
        	for(int i=0;i<9;i++) { 
        		rows[i]=new HashMap<>();
        		colums[i]=new HashMap<>();
        		boxes[i]= new HashMap<>();
        	}
        	
    // 先遍历每一行,在遍历每一列
        	for(int row=0;row<9;row++){ 
    // 遍历每行每一列
        		for(int colum=0;colum<9;colum++) { 
    // 或去row行colum列的值
        			char num=board[row][colum];
    // 如果不是.号就转化为数字存入哈希中
        			if (num!='.') { 
    					int n=(int)num;
    // 存入每一行
    					rows[row].put(n, rows[row].getOrDefault(n, 0)+1);
    // 存入每一列
    					colums[colum].put(n, colums[colum].getOrDefault(n, 0)+1);
    // 设定每一个小独数，跟人工智能的
    					int box_index=(row/3)*3+(colum/3);
    // 存入每一个小独数的元素
    					boxes[box_index].put(n, boxes[box_index].getOrDefault(n, 0)+1);
    // 在判断一下这三个哈希表的值是否存在大于1的，因为大于一就说明有重复的值
    					if (rows[row].get(n)>1||colums[colum].get(n)>1||boxes[box_index].get(n)>1) { 
    						return false;
    					}
    				}
        		}
        	}
        	
        	
        	return true;
        }
    }

> 本人csdn博客：https://blog.csdn.net/weixin\_46654114
> 
> 转载说明：跟我说明，务必注明来源，附带本人博客连接。

[image-20210902204936105]: /images/20220828/3dad584a03454bbd8b165c491de62951.png
[36.]: https://leetcode-cn.com/problems/valid-sudoku/
[img]: /images/20220828/3ec2ab698e40492cbca2da76564a9f9c.png