leecode刷题（9）-- 有效的数独
leecode刷题（9）-- 有效的数独
希希里之海
发布于 2019-01-07 13:59:53
发布于 2019-01-07 13:59:53
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。
- 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。
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。
思路：
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。
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原始发表：2019-01-03 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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