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LeetCode 36. 有效的数独

难度：$middle$

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题目描述

请你判断一个 $9x9$ 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。

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 ${}^{\prime }{.}^{\prime }$ 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 ${}^{\prime }{.}^{\prime }$

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算法

(暴力枚举)

• 判断每一行是否存在相同的数字。
• 判断每一列是否存在相同的数字。
• 判断每一个单独的小方格是否存在相同的数字。

复杂度分析

• 时间复杂度：$O(1)$，数独共有 81 个单元格，只需要对每个单元格遍历即可。

• 空间复杂度 : $O(1)$，只需要一个判重数组即可。

C++ 代码

class Solution {
public:bool isValidSudoku(vector<vector<char>>& board) {bool st[9];//判断行for (int i = 0; i < 9; i ++) {memset(st, 0, sizeof st);for (int j = 0; j < 9; j ++) {if (board[i][j] != '.') {int t = board[i][j] - '1';if (st[t]) return false;st[t] = true;}}}//判断列for (int i = 0; i < 9; i ++) {memset(st, 0, sizeof st);for (int j = 0; j < 9; j ++) {if (board[j][i] != '.') {int t = board[j][i] - '1';if (st[t]) return false;st[t] = true;}}}//判断小方格for (int i = 0; i < 9; i += 3) {for (int j = 0; j < 9; j += 3) {memset(st, 0, sizeof st);for (int x = 0; x < 3; x ++) {for (int y = 0; y < 3; y ++) {if (board[i + x][j + y] != '.') {int t = board[i + x][j + y] - '1';if (st[t]) return false;st[t] = true;}}}}}return true;}
};

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