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LeetCode 36. 有效的数独 - 解题思路与实现详解

news 2025/9/1 14:48:47

LeetCode 36. 有效的数独 - 解题思路与实现详解

题目描述

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：

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。

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。

解题思路分析

1. 问题理解

数独验证的核心是检查三个维度：

行检查：每一行中的数字1-9不能重复
列检查：每一列中的数字1-9不能重复
3x3宫格检查：每个3x3的小宫格中数字1-9不能重复

2. 关键思考点

2.1 如何处理空白格（‘.’）

空白格用’.'表示，在验证时需要考虑：

集合去重时会自动忽略’.’
需要正确计算有效数字的数量

2.2 3x3宫格的遍历

如何正确遍历9个3x3宫格是关键：

每个宫格的起始位置：(0,0), (0,3), (0,6), (3,0), (3,3), (3,6), (6,0), (6,3), (6,6)
使用双重循环：for x in range(0, 9, 3): for y in range(0, 9, 3)

2.3 列数据的提取

Python中不能直接用board[:][y]获取列，需要使用列表推导式：

column_y = [row[y] for row in board]

代码实现

方法一：集合验证法

class Solution:def isValidSudoku(self, board: list[list[str]]) -> bool:# 检查3x3宫格for x in range(0, 9, 3):for y in range(0, 9, 3):# 提取当前3x3宫格的所有元素grids = [board[x][y], board[x][y+1], board[x][y+2],board[x+1][y], board[x+1][y+1], board[x+1][y+2],board[x+2][y], board[x+2][y+1], board[x+2][y+2]]# 验证：去重后的数字个数 + 空白格个数 = 9if (len(set(grids)) + grids.count('.')) - 1 != 9:return False# 检查每一行for x in range(9):if (len(set(board[x])) + board[x].count('.')) - 1 != 9:return False# 检查每一列for y in range(9):column_y = [row[y] for row in board]if (len(set(column_y)) + column_y.count('.')) - 1 != 9:return Falsereturn True

方法二：哈希表优化法

class Solution:def isValidSudoku(self, board: list[list[str]]) -> bool:# 使用哈希表记录每行、每列、每个宫格中数字的出现情况rows = [set() for _ in range(9)]cols = [set() for _ in range(9)]boxes = [set() for _ in range(9)]for i in range(9):for j in range(9):num = board[i][j]if num == '.':continue# 检查行if num in rows[i]:return Falserows[i].add(num)# 检查列if num in cols[j]:return Falsecols[j].add(num)# 检查3x3宫格box_index = (i // 3) * 3 + j // 3if num in boxes[box_index]:return Falseboxes[box_index].add(num)return True

关键知识点

1. Python集合操作

# 集合去重
numbers = [1, 2, 2, 3, 4, 4, 5]
unique_numbers = set(numbers)  # {1, 2, 3, 4, 5}# 集合长度
len(unique_numbers)  # 5

2. 列表推导式

# 提取矩阵的列
matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
column_0 = [row[0] for row in matrix]  # [1, 4, 7]
column_1 = [row[1] for row in matrix]  # [2, 5, 8]

3. 3x3宫格索引计算

# 计算(i,j)位置属于哪个3x3宫格
box_index = (i // 3) * 3 + j // 3# 宫格编号对应关系：
# 0 1 2
# 3 4 5  
# 6 7 8

解题过程中的常见错误

1. 错误理解集合去重

# 错误示例
grids = ['.', '1', '2', '.', '3', '4', '5', '6', '7']
unique_count = len(set(grids))  # 8 (包含了'.')
# 正确做法
valid_count = len(set(grids)) + grids.count('.') - 1  # 7

2. 错误的列提取方式

# 错误示例
board = [[1,2,3], [4,5,6], [7,8,9]]
column = board[:][0]  # 这样不会得到第一列# 正确做法
column = [row[0] for row in board]  # [1, 4, 7]

3. 3x3宫格遍历错误

# 错误示例：没有在循环内重新计算grids
grids = [board[0][0], board[0][1], ...]  # 只计算了一次
for x in range(0, 9, 3):for y in range(0, 9, 3):# 应该在这里重新计算grids

复杂度分析

时间复杂度

方法一：O(n²)，其中n=9，需要遍历整个数独板
方法二：O(n²)，但实际运行更快，因为使用哈希表查找

空间复杂度

方法一：O(1)，只使用常数空间
方法二：O(n)，需要存储行、列、宫格的哈希表

测试用例

# 测试用例1：有效数独
board1 = [["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"]
]
# 预期结果：True# 测试用例2：无效数独（第一行有重复）
board2 = [["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"]
]
# 预期结果：False