LeetCode【0036】有效的数独

本文目录

1 中文题目
2 求解方法：python内置函数set
- 2.1 方法思路
- 2.2 Python代码
- 2.3 复杂度分析
3 题目总结

1 中文题目

请根据以下规则判断一个 9 x 9 的数独是否有效。

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。（见下方的参考示例图）

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。
空白格用 ‘.’ 表示。

示例：

在这里插入图片描述

对于上面的数独，其输入格式如下：
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

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（1-9）或者 ‘.’

2 求解方法：python内置函数set

2.1 方法思路

方法核心

使用三组集合分别记录每行、每列和每个3x3方块中已出现的数字
一次遍历完成所有验证
使用数学公式计算粗实线分割出来的3x3方块的索引

实现步骤

（1）初始化数据结构：

创建9个集合用于存储每行的数字
创建9个集合用于存储每列的数字
创建9个集合用于存储每个3x3方块的数字

（2）遍历数独板：

双层循环遍历9x9的数独板
对于每个位置，获取当前数字
跳过空格（用’.'表示）
计算当前位置所在的3x3方块的索引
检查数字是否重复出现
将不重复的数字添加到对应集合中

（3）验证过程：

检查当前数字是否已在当前行出现
检查当前数字是否已在当前列出现
检查当前数字是否已在当前3x3方块出现
如果出现重复，立即返回False
如果遍历完成没有重复，返回True

方法示例

输入数独板示例（部分）：
[["5","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".",".","."],[".","9","8",".",".",".",".","6","."],...
]详细执行过程：1. 初始化：rows = [set(), set(), set(), ...] (9个空集合)cols = [set(), set(), set(), ...] (9个空集合)boxes = [set(), set(), set(), ...] (9个空集合)2. 处理第一个位置 (0,0)：- 数字为 "5"- box_index = (0 // 3) * 3 + (0 // 3) = 0- 检查 rows[0], cols[0], boxes[0] 都不包含 "5"- 将 "5" 添加到这三个集合中3. 处理第二个位置 (0,1)：- 数字为 "3"- box_index = (0 // 3) * 3 + (1 // 3) = 0- 检查集合，添加数字4. 继续处理每个位置...示例中3x3方块索引的计算：
- 位置(0,0): (0//3)*3 + 0//3 = 0
- 位置(0,4): (0//3)*3 + 4//3 = 1
- 位置(4,4): (4//3)*3 + 4//3 = 4

2.2 Python代码

class Solution:def isValidSudoku(self, board: List[List[str]]) -> bool:# 初始化用于存储每行、每列和每个3x3方块中数字出现情况的集合rows = [set() for _ in range(9)]cols = [set() for _ in range(9)]boxes = [set() for _ in range(9)]# 遍历整个数独板for i in range(9):for j in range(9):# 获取当前位置的数字num = board[i][j]# 如果是空格，继续下一个位置if num == '.':continue# 计算当前位置所在的3x3方块的索引# box_index = (行号 // 3) * 3 + (列号 // 3)box_index = (i // 3) * 3 + j // 3# 检查当前数字是否已经在对应的行、列或3x3方块中出现过if (num in rows[i] or num in cols[j] or num in boxes[box_index]):return False# 将当前数字添加到对应的集合中rows[i].add(num)cols[j].add(num)boxes[box_index].add(num)# 遍历完成且没有发现重复，返回Truereturn True