36. 有效的数独（Valid Sudoku）题解（C语言）

题目描述

判断一个 9×9 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 3×3 宫内只能出现一次。

注意：

• 空白格用 '.' 表示；

• 一个有效的数独不一定是可解的，只需要验证已经填入的数字是否有效即可。

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：左上角 3×3 宫格内有两个 '8'，因此无效。

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解题思路

本题的核心是 检查三种约束条件：

1. 每行数字不能重复；

2. 每列数字不能重复；

3. 每个 3×3 宫格内数字不能重复。

方法

• 使用三个二维数组来记录数字是否出现过：

  • row[9][9]：行出现记录

  • col[9][9]：列出现记录

  • box[9][9]：3×3 宫格出现记录

• 遍历整个棋盘：

  1. 如果当前格是 '.'，跳过；

  2. 否则将字符转为数字索引 num = board[i][j] - '1'；

  3. 计算当前格所属的宫格索引 boxIndex = (i / 3) * 3 + j / 3；

  4. 如果 row[i][num]、col[j][num] 或 box[boxIndex][num] 已经标记过，则返回 false；

  5. 否则将其标记为已出现。

• 遍历完成后仍未发现冲突，则数独有效，返回 true。

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C语言实现

#include <stdbool.h>

bool isValidSudoku(char** board, int boardSize, int* boardColSize) {
    int row[9][9] = {0};
    int col[9][9] = {0};
    int box[9][9] = {0};

    for(int i = 0; i < 9; i++){
        for(int j = 0; j < 9; j++){
            if(board[i][j] == '.') continue;

            int num = board[i][j] - '1';       // 数字转换为索引 0~8
            int boxIndex = (i / 3) * 3 + j / 3; // 计算 3x3 宫格索引

            if(row[i][num] || col[j][num] || box[boxIndex][num])
                return false;

            row[i][num] = 1;
            col[j][num] = 1;
            box[boxIndex][num] = 1;
        }
    }

    return true;
}

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算法分析

• 时间复杂度：O(9×9) = O(1)，固定大小棋盘，遍历每个格子一次；

• 空间复杂度：O(9×9) = O(1)，使用固定大小的三个二维数组存储状态。

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总结

本题属于 经典的状态记录题，核心技巧是：

1. 使用辅助数组记录行、列、宫格的状态；

2. 宫格索引计算公式：boxIndex = (i/3)*3 + j/3；

3. 遇到 '.' 跳过即可。

面试时，如果能讲出这一思路，并且写出简洁的 C 语言实现，基本可以轻松通过。

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