题目

Determine if a 9 x 9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

1.Each row must contain the digits 1-9 without repetition.

2.Each column must contain the digits 1-9 without repetition.

3.Each of the nine 3 x 3 sub-boxes of the grid must contain the digits 1-9 without repetition.

Note:

  • A Sudoku board (partially filled) could be valid but is not necessarily solvable.
  • Only the filled cells need to be validated according to the mentioned rules.

Example 1:

Input: 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"]]
Output: true

Example 2:

Input: 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"]]
Output: false
Explanation: Same as Example 1, except with the 5 in the top left corner being modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.

Constraints:

  • board.length == 9
  • board[i].length == 9
  • board[i][j] is a digit 1-9 or '.'.

思路

方法1 (Java)

设置一个Set集合,遍历数独表二维数组,将当前遍历到的位置的元素以3个字符串形式记录下它所在的行,列和九宫格信息,并加入Set集合。若Set集合中存在某一元素的任意行/列/九宫格信息,则认为该表不符合规则。

设当前元素值为val,位置为(i, j),位于第i/3*3+j/3个九宫格(从0开始),则该元素的信息可以设置为:

行信息:"val in row i"

列信息:"val in col j"

九宫格信息:"val in box i/3*3+j/3"

class Solution {
public boolean isValidSudoku(char[][] board) {
HashSet<String> set = new HashSet<>();
for(int i = 0; i < 9; i++){
for(int j = 0; j < 9; j++){
if(board[i][j] == '.') continue;
String row = board[i][j] + " in row " + i;
String col = board[i][j] + " in col " + j;
String box = board[i][j] + " in box " + (i/3*3+j/3);
if(set.contains(row) || set.contains(col) || set.contains(box)) return false;
set.add(row);
set.add(col);
set.add(box);
}
}
return true;
}
}

方法2(Java)

0-8,对每一个i,设置3个boolean数组分别检测第i行、第i列和第i个九宫格,数组的index代表被检测的值。

对于每个i的值,j0-8意味着检测:

i行, 第j个数是否已位于行i中;

i列,第j个数是否已位于列i中;

i个九宫格中,该九宫格从左上角到右下角的第j数是否已位于该九宫格中。

 

设置3个数组,

boolean[] checkRow 检查位置(i, j)的值是否已存在第i行

boolean[] checkCol 检查位置(j, i)的值是否已存在第i列

boolean[] checkBox 检查位置(i/3*3+j/3, j%3*3+j%3)的值是否已存在第i个九宫格中

例如当i为4,j从0-8,对应行、列、九宫格的遍历过程

  0 1 2   3 4 5   6 7 8
0 o o o | o 1 o | o o o
1 o o o | o 2 o | o o o
2 o o o | o 3 o | o o o
------------------------
3 o o o | o 4 o | o o o
4 1 2 3 | 4 5 6 | 7 8 9
5 o o o | o 6 o | o o o
------------------------
6 o o o | o 7 o | o o o
7 o o o | o 8 o | o o o
8 o o o | o 9 o | o o o 第4个九宫格中
1 2 3
4 5 6
7 8 9
class Solution {
public boolean isValidSudoku(char[][] board) {
for(int i = 0; i < 9; i++){
boolean[] checkRow = new boolean[9];
boolean[] checkCol = new boolean[9];
boolean[] checkBox = new boolean[9];
for(int j = 0; j < 9; j++){
if(board[i][j] == '.') {}
else if(checkRow[board[i][j]-'1']) return false;
else checkRow[board[i][j]-'1'] = true; if(board[j][i] == '.'){}
else if(checkCol[board[j][i]-'1']) return false;
else checkCol[board[j][i]-'1'] = true; int m = i/3*3+j/3, n = i%3*3+j%3;
if(board[m][n] == '.'){}
else if(checkBox[board[m][n]-'1']) return false;
else checkBox[board[m][n]-'1'] = true;
}
}
return true;
}
}

LeetCode 36. Valid Sudoku (Medium)的更多相关文章

  1. LeetCode:36. Valid Sudoku,数独是否有效

    LeetCode:36. Valid Sudoku,数独是否有效 : 题目: LeetCode:36. Valid Sudoku 描述: Determine if a Sudoku is valid, ...

  2. 蜗牛慢慢爬 LeetCode 36.Valid Sudoku [Difficulty: Medium]

    题目 Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules. The Sudoku board could ...

  3. LeetCode 36 Valid Sudoku

    Problem: Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules. The Sudoku board ...

  4. Java [leetcode 36]Valid Sudoku

    题目描述: Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules. The Sudoku board cou ...

  5. 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 bo ...

  6. [LeetCode] 36. Valid Sudoku 验证数独

    Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to th ...

  7. [leetcode]36. Valid Sudoku验证数独

    Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to th ...

  8. LeetCode 36. Valid Sudoku (C++)

    题目: Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according t ...

  9. LeetCode 36 Valid Sudoku(合法的数独)

    题目链接: https://leetcode.com/problems/valid-sudoku/?tab=Description   给出一个二维数组,数组大小为数独的大小,即9*9  其中,未填入 ...

随机推荐

  1. laravel DB 类库

    DB 类操作数据库    基本用法: DB::table('tableName'); 获取操作tableName 表        增加信息        对数据库中的某个表增加数据主要有两个函数可以 ...

  2. 从零搭建vsftpd

    先吐槽一下这个工具,配置繁琐,限制规则复杂,报错信息不够详细,学起来吃力. 准备工作 [root@vsftp-server ~]# mkdir /data/ #创建ftp目录 [root@vsftp- ...

  3. [bzoj1741]穿越小行星群

    将每一行/每一列作为一个点,对于一个障碍(x,y),要么第x行和第y列的状态(是否攻击)只需要有一个就可以了,将第x行和第y列连边,就是二分图的最小点覆盖=最大匹配数. 1 #include<b ...

  4. [atARC084D]Small Multiple

    构造一张图:$\forall x$,向$10x$连一条边权为0的边,向$x+1$连1条边权为1的边,那么0到$i$的代价即为$i$各位数字之和 考虑到我们只关心于当前点的两个特征:1.模$n$的余数( ...

  5. [luogu5423]Valleys

    先考虑不要求有洞,那么可以将所有权值排序,然后不断插入,那么一个连通块就是一个答案,加上连通块大小即可考虑并查集如何判断是否有洞,可以发现对于任意一个无洞的直角多边形,都有$90度内角-90度外角=4 ...

  6. python中使用正则表达式处理文本(仅记录常用方法和参数)

    标准库模块 python中通过re模块使用正则表达式 import re 常用方法 生成正则表达式对象 compile(pattern[,flags]) pattern:正则表达式字符串 flags: ...

  7. javascript-初级-day05js函数传参

    JS基础-01 函数传参.参数类型-1 <!DOCTYPE HTML> <html> <head> <meta http-equiv="Conten ...

  8. Collections集合工具类的常用方法

    Collections集合工具类的方法 addAll与shuffle import java.util.ArrayList; import java.util.Collections; /* - ja ...

  9. ARC 119 补题记录

    这把感觉质量很高. \(E\) \(E\)比较简单所以先写个\(E\),考虑就一个置换操作来说改变的只有两端的值. 考虑\(|a_i - a_{i - 1}|\)变成区间,则我们考虑分类讨论,发现只有 ...

  10. LOJ #6044 -「雅礼集训 2017 Day8」共(矩阵树定理+手推行列式)

    题面传送门 一道代码让你觉得它是道给初学者做的题,然鹅我竟没想到? 首先考虑做一步转化,我们考虑将整棵树按深度奇偶性转化为一张二分图,即将深度为奇数的点视作二分图的左部,深度为偶数的点视作二分图的右部 ...