36. 有效的数独【 力扣(LeetCode) 】

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

1. 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。
2. 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;
    }
};