wordpress 使用浏览器缓存seo基础入门免费教程
一、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。
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。
二、测试用例
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 '.'
三、解题思路
- 基本思路：
一力破万法，检查是否满足数独的三个条件就可以了。 - 具体思路：一次遍历就可以检查三个条件，就是需要一些技巧。
- 行唯一：判断每一行中出现的数字是否唯一 【正常遍历】
- 列唯一：判断每一列中出现的数字是否唯一 【行列交换】
- 九宫格唯一：判断每一个九宫格中出现的数字是否唯一 【特殊映射】
四、参考代码
时间复杂度： O ( 1 ) \Omicron(1) O(1)【数独是固定大小的，所以都是常数级复杂度】
空间复杂度： O ( 1 ) \Omicron(1) O(1)【数独是固定大小的，所以都是常数级复杂度】
class Solution {
public:bool isValidSudoku(vector<vector<char>>& board) {int n = board.size(), m = board[0].size();for (int i = 0; i < n; i++) {vector<vector<bool>> num(3, vector<bool>(m, false));for (int j = 0; j < m; j++) { // 行唯一if (board[i][j] != '.') {if (num[0][board[i][j] - '1']) {return false;} else {num[0][board[i][j] - '1'] = true;}}if (board[j][i] != '.') { // 列唯一if (num[1][board[j][i] - '1']) {return false;} else {num[1][board[j][i] - '1'] = true;}}int r = i / 3 * 3 + j / 3, c = (i % 3) * 3 + j % 3;if (board[r][c] != '.') { // 九宫格唯一if (num[2][board[r][c] - '1']) {return false;} else {num[2][board[r][c] - '1'] = true;}}}}return true;}
};