LeetCode 题解工作台

有效的数独

请你判断一个 9 x 9 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。(请参考示例图) 注意: 一个有效的数独(部分已被填充)不一…

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中等 · 数组·哈希·扫描

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答案摘要

有效的数独满足以下三个条件: - 每一行中的数字都不重复;

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题目描述

请你判断一个 9 x 9 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)

 

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 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(1-9)或者 '.'
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解题思路

方法一:一次遍历

有效的数独满足以下三个条件:

  • 每一行中的数字都不重复;
  • 每一列中的数字都不重复;
  • 每一个 3×33 \times 3 的宫格中的数字都不重复。

遍历数独,对于每个数字,判断其所在的行、列 以及 3×33 \times 3 的宫格是否已经出现过该数字,如果是,则返回 false。遍历结束,返回 true

时间复杂度 O(C)O(C),空间复杂度 O(C)O(C),其中 CC 是数独中的空格数。本题中 C=81C=81

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class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        row = [[False] * 9 for _ in range(9)]
        col = [[False] * 9 for _ in range(9)]
        sub = [[False] * 9 for _ in range(9)]
        for i in range(9):
            for j in range(9):
                c = board[i][j]
                if c == '.':
                    continue
                num = int(c) - 1
                k = i // 3 * 3 + j // 3
                if row[i][num] or col[j][num] or sub[k][num]:
                    return False
                row[i][num] = True
                col[j][num] = True
                sub[k][num] = True
        return True
speed

复杂度分析

指标
时间Depends on the final approach
空间Depends on the final approach
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面试官常问的追问

外企场景
  • question_mark

    Do you recognize how duplicate checks in rows, columns, and sub-boxes can be optimized using hash sets?

  • question_mark

    Can you map each board cell to its 3x3 sub-box index without extra loops?

  • question_mark

    Will you handle empty cells efficiently to avoid unnecessary hash operations?

warning

常见陷阱

外企场景
  • error

    Confusing the 3x3 sub-box indexing, which can lead to missed duplicates or false positives.

  • error

    Checking only rows and columns but forgetting sub-boxes, which breaks Sudoku rules.

  • error

    Attempting to scan sub-boxes with nested loops for every cell instead of using precomputed hash sets, causing redundant computations.

swap_horiz

进阶变体

外企场景
  • arrow_right_alt

    Determine the minimum changes required to make a partially filled Sudoku board valid.

  • arrow_right_alt

    Validate a Sudoku board of arbitrary size N x N with sqrt(N) x sqrt(N) sub-boxes.

  • arrow_right_alt

    Implement a Sudoku solver that fills empty cells while maintaining validity at each step.

help

常见问题

外企场景

有效的数独题解:数组·哈希·扫描 | LeetCode #36 中等