Leetcode刷题100天—36. 有效的数独(数组+哈希)—day26

Posted 2021-10-10 神的孩子都在歌唱

前言：

作者：神的孩子在歌唱

大家好，我叫运智

36. 有效的数独

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：

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

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