leecode刷题（9）-- 有效的数独

leecode刷题（9）-- 有效的数独

有效的数独

描述：

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。

上图是一个部分填充的有效的数独。

数独部分空格内已填入了数字，空白格用 ‘.‘ 表示。

示例 1:

输入:
[
  ["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:
[
  ["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。

说明:

• 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。
• 给定数独序列只包含数字 1-9 和字符 ‘.‘ 。
• 给定数独永远是 9x9 形式的。

这道题，其实我真的不会。。。虽然知道是依次检查行、检查列、检查 9 个 3 X 3 的小九宫格是否出现重复元素，如果出现返回 false，否则返回 true。但我不知道怎么写，特别是关于判断小九宫格中每个格点坐标的值，不知道应该如何去编写代码，我笨（敲自己两下）。所以在网上查找了一下关于这道问题各位大神的解，其实自己看的并不是很懂，这里在此记录一下，希望自己以后能力提高后能独立写出来，加油！

代码如下：

public boolean isValidSudoku(char[][] board) {
        for (int i = 0; i < 9; i++) {
            HashSet<Character> row = new HashSet<>();
            HashSet<Character> column = new HashSet<>();
            HashSet<Character> cube = new HashSet<>();
            for (int j = 0; j < 9; j++) {
                // 检查第i行，在横坐标位置
                if (board[i][j] != ‘.‘ && !row.add(board[i][j]))
                    return false;
                // 检查第i列，在纵坐标位置
                if (board[j][i] != ‘.‘ && !column.add(board[j][i]))
                    return false;
                // 行号+偏移量
                int RowIndex = 3 * (i / 3) + j / 3;
                // 列号+偏移量
                int ColIndex = 3 * (i % 3) + j % 3;
                //每个小九宫格，总共9个
                if (board[RowIndex][ColIndex] != ‘.‘
                        && !cube.add(board[RowIndex][ColIndex]))
                    return false;
            }
        }
        return true;
    }

第 i 个九宫格的第 j 个格点的行号可表示为 i/3*3+j/3

第 i 个九宫格的第 j 个格点的列号可表示为 i%3*3+j%3。

原文地址：https://www.cnblogs.com/weixuqin/p/10215952.html