题目

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：
[中等] 36. 有效的数独 - 图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。

解答 & 代码

解法一：暴力搜索

对每个数字，分别验证行、列、3*3 宫中是否存在重复数字

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        int len = 9;
        // 遍历数独的每个位置，验证已填入的数字是否有效
        for(int row = 0; row < len; ++row)
        {
            for(int col = 0; col < len; ++col)
            {
                // 如果当前位置是空的，则跳过
                if(board[row][col] == '.')
                    continue;

                // 判断行是否存在重复数字
                for(int j = col + 1; j < len; ++j)
                {
                    if(board[row][j] == board[row][col])
                        return false;
                }
                // 判断列是否存在重复数字
                for(int i = row + 1; i < len; ++i)
                {
                    if(board[i][col] == board[row][col])
                        return false;
                }
                // 判断 3*3 的宫内是否存在重复数字
                for(int i = row / 3 * 3; i < (row / 3 + 1) * 3; ++i)
                {
                    for(int j = col / 3 * 3; j < (col / 3 + 1) * 3; ++j)
                    {
                        if(i == row && j == col)
                            continue;
                        if(board[i][j] == board[row][col])
                            return false;
                    }
                }
            }
        }
        return true;
    }
};

执行结果：

执行结果：通过

执行用时：20 ms, 在所有 C++ 提交中击败了 51.56% 的用户
内存消耗：17.4 MB, 在所有 C++ 提交中击败了 83.09% 的用户

解法二：一次遍历（哈希表）

设置哈希表，分别存储每行、每列、每个 3*3 格子中各个数字出现的次数。
由于只有 9 个数字，因此可以直接用数组代替哈希表

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        // 存储每行中每个数字出现的次数
        vector<vector<int>> rowsMap(9, vector<int>(9, 0));
        // 存储每列中每个数字出现的次数
        vector<vector<int>> colsMap(9, vector<int>(9, 0));
        // 存储每个 3*3 格子中每个数字出现的次数
        vector<vector<vector<int>>> boxsMap(3, vector<vector<int>>(3, vector<int>(9, 0)));

        // 遍历数独
        for(int row = 0; row < 9; ++row)
        {
            for(int col = 0; col < 9; ++col)
            {
                // 跳过非数字的空格
                if(board[row][col] == '.')
                    continue;

                // 如果当前数字在哈希表中已存在，则直接返回 false
                int num = board[row][col] - '0' - 1;
                if(rowsMap[row][num] == 1 || colsMap[col][num] == 1 ||
                    boxsMap[row / 3][col / 3][num] == 1)
                    return false;
                else    // 否则，哈希表对应位置 + 1
                {
                    ++rowsMap[row][num];
                    ++colsMap[col][num];
                    ++boxsMap[row / 3][col / 3][num];
                }
            }
        }
        return true;
    }
};

执行结果：

执行结果：通过

执行用时：12 ms, 在所有 C++ 提交中击败了 94.00% 的用户
内存消耗：19.6 MB, 在所有 C++ 提交中击败了 11.57% 的用户