LeetCode【0036】有效的数独

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：

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True
        return True

2.3 复杂度分析

• 时间复杂度：O(1)
  • 固定大小的9x9网格
  • 遍历一次数独数组
  • 每次检查和添加操作都是O(1)
  • 总操作次数是常数
• 空间复杂度：O(1)
  • 使用固定大小的集合
  • 9个集合用于行
  • 9个集合用于列
  • 9个集合用于3x3方块

---

3 题目总结

题目难度：中等
数据结构：二维数组
应用算法：Python内置函数set

原文地址：https://blog.csdn.net/qq_32882309/article/details/143741003

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