36. 有效的数独
36. 有效的数独
张伦聪zhangluncong
发布于 2022-10-26 18:08:26
发布于 2022-10-26 18:08:26
代码可运行
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代码可运行
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。
解1： 掌握核心科技,不过核心科技太难掌握。下面公式不知道哪个大神推导出来的，非常难。看解2。
x = (i / 3) * 3 + (k % 9) / 3;
y = j % 3 + ((k / 9) % 3) * 3;public boolean isValidSudoku(char[][] board) {
Set<Character> rowSet = new HashSet<>();
Set<Character> lineSet = new HashSet<>();
Set<Character> sudokuSet = new HashSet<>();
int k = 0;
int x = 0;
int y = 0;
for (int i = 0; i < 9; i++) {
rowSet.clear();
lineSet.clear();
sudokuSet.clear();
for (int j = 0; j < 9; j++) {
//行
if (board[i][j] != '.' && !rowSet.add(board[i][j])) {
return false;
}
//列
if (board[j][i] != '.' && !lineSet.add(board[j][i])) {
return false;
}
x = (i / 3) * 3 + (k % 9) / 3;
y = j % 3 + ((k / 9) % 3) * 3;
//小格子
if (board[x][y] != '.' && !sudokuSet.add(board[x][y])) {
return false;
}
k += 1;
}
}
return true;
}解2：牺牲了一点时间，比较容易理解。
public boolean isValidSudoku(char[][] board) {
Set<Character> rowSet = new HashSet<>();
Set<Character> lineSet = new HashSet<>();
Set<Character> sudokuSet = new HashSet<>();
int m = board.length;
int n = board[0].length;
//行列
for (int i = 0; i < m; i++) {
rowSet.clear();
lineSet.clear();
for (int j = 0; j < n; j++) {
//行
if (board[i][j] != '.' && !rowSet.add(board[i][j])) {
return false;
}
//列
if (board[j][i] != '.' && !lineSet.add(board[j][i])) {
return false;
}
}
}
//小格子
for (int i = 0; i < m - 2; i = i + 3) {
for (int j = 0; j < n - 2; j = j + 3) {
//外面两个循环确定了小格子左上角第一个元素
sudokuSet.clear();
for (int a = i; a < i + 3; a++) {
for (int b = j; b < j + 3; b++) {
if (board[a][b] != '.' && !sudokuSet.add(board[a][b])) {
return false;
}
}
}
}
}
return true;
}本文参与 腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。
原始发表：2018-07-24，如有侵权请联系 cloudcommunity@tencent.com 删除
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