leetcode-有效的数独

这是我参与8月更文挑战的第25天，活动详情查看：8月更文挑战

今晚还是有乒乓球赛，跟昨天一样，下午抽空先做下leetcode的题目。昨天比赛赢多输少，不过还是被淘汰了，没有把握住关键场次的关键球，有点可惜。打算比赛后好好买块乒乓板然后每月去打个3-4次，以这种方式重燃了小时候去国球的热情，也是意料之外。
今天做的是leetcode的第36题。

题目

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。

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）
数独部分空格内已填入了数字，空白格用 '.' 表示。

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。

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。

思路

不太明白leetcode的难度定义逻辑，这题应该挺水的，竟然是个中等。
当然前提是要了解数独的规则，简单来说，一个99的大区域，有9行、9列、9个33的小方块，每个行、列、小方块就是1个小区域，每个小区域中，数字1-9各出现1次。
行和列都很好理解，小方块的画个图：

本题简单的地方在于，只要判断当前已经存在数字的地方，是否符合规则，而不用判断这个数独是否有解。
那我们直接定义3个9*9布尔类型的二维数组line、column、box，分别代表行、列、小方块；
line的第1维代表数组的0-8行，第2维代表1-9这9个数字是否出现过
column的第1维代表数组的0-8列，第2维代表1-9这9个数字是否出现过
box的第1维代表0-8个小方框，第2维代表1-9这9个数字是否出现过
默认值都是false，如果数组在这个位置出现了，就把它改成true，如果发现位置上已经是true了，就代表在1个小区域内出现了第2次，那就是无效的数独了。
可能存在的难点在于怎么判断1个单元格属于哪个小方块，可以这样：boxIndex = (i/3)*3 + j/3

Java版本代码

class Solution {
    public boolean isValidSudoku(char[][] board) {
        // 定义3个2维数据，分别代表 行、列、框
        // line的第1维代表数组的0-8行，第2维代表1-9这9个数字是否出现过
        // column的第1维代表数组的0-8列，第2维代表1-9这9个数字是否出现过
        // box的第1维代表0-8个小方框，第2维代表1-9这9个数字是否出现过
        // 初始都是false
        boolean[][] line = new boolean[9][9];
        boolean[][] column = new boolean[9][9];
        boolean[][] box = new boolean[9][9];
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                char c = board[i][j];
                if (c != '.') {
                    int val = c - '0';
                    if (line[i][val - 1] || column[j][val - 1] || box[(i/3)*3 + j/3][val - 1]) {
                        return false;
                    } else {
                        line[i][val - 1] = true;
                        column[j][val - 1] = true;
                        box[(i/3)*3 + j/3][val - 1] = true;
                    }
                }
            }
        }
        return true;
    }
}