【刷题记录】36. 有效的数独Offer 驾到，掘友接招！我正在参与2022春招打卡活动，点击查看活动详情。模拟判断数

Offer 驾到，掘友接招！我正在参与2022春招打卡活动，点击查看活动详情。

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。

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 '.'

二丶思路分析

数组

这道题目是判断数独的有效性 所以我们可以定义三个二维数组

rows 记录每行数字出现的情况
colnum 记录每列数字出现的情况
area 记录每个方块数字出现的情况
我们可以推出方块编号和行列的关系为 idx = i / 3 * 3 + j / 3
遍历 board，如果违反数独的规则，则返回 false,结束都没违反则返回true

三、代码实现

class Solution {
    public boolean isValidSudoku(char[][] board) {
        
        //记录每行，列 ，每个方块出现数字的情况
        boolean[][] row = new boolean[9][9];
        boolean[][] colnum = new boolean[9][9];
        boolean[][] area = new boolean[9][9];

        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                if (board[i][j] == '.') continue;
                int c = board[i][j] - '1';
                int idx = i / 3 * 3 + j / 3;
                if (!row[i][c] && !colnum[j][c] && !area[idx][c]) {
                    //标记为出现过
                    row[i][c] = colnum[j][c] = area[idx][c] = true;
                } else {
                    return false;
                }
            }
        }
        return true;
    }
}

复杂度分析

时间复杂度： $O(1)$
空间复杂度： $O(1)$

运行结果

总结

这道题目就是一道按照 数独的规则 进行模拟的题目，然后遍历，看是否完全满足。

其中还有一个就是找出每个方块编号和行列的对于关系，或者直接用一个三维数组来代替也行。

继续加油~~