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。
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:
由于只要我们判断是否为有效的数独。
所以我们只需要对 board
中出现的数进行判断,如果 board
中有数违反了数独的规则,返回 false
,否则返回 true
。
直观上,我们很容易想到使用哈希表来记录某行/某列/某个小方块出现过哪些数字,来帮助我们判断是否符合「有效数独」的定义。
这道题唯一的难点可能是在于如何确定某个数落在哪个小方块中,我们可以去小方块进行编号:
然后推导出小方块编号和行列的关系为:idx = i / 3 * 3 + j / 3
。
class Solution {
public boolean isValidSudoku(char[][] board) {
Map<Integer, Set<Integer>> row = new HashMap<>(), col = new HashMap<>(), area = new HashMap<>();
for (int i = 0; i < 9; i++) {
row.put(i, new HashSet<>());
col.put(i, new HashSet<>());
area.put(i, new HashSet<>());
}
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.get(i).contains(c) && !col.get(j).contains(c) && !area.get(idx).contains(c)) {
row.get(i).add(c);
col.get(j).add(c);
area.get(idx).add(c);
} else {
return false;
}
}
}
return true;
}
}
9*9
的棋盘里,复杂度不随数据变化而变化。复杂度为9*9
的棋盘里,复杂度不随数据变化而变化。复杂度为大多数的哈希表计数问题,都能转换为使用数组解决。
虽然时间复杂度一样,但是哈希表的更新和查询复杂度是平均为
,而定长数组的的更新和查询复杂度是严格
。
因此从执行效率上来说,数组要比哈希表快上不少:
class Solution {
public boolean isValidSudoku(char[][] board) {
boolean[][] row = new boolean[9][9], col = new boolean[9][9], 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] && !col[j][c] && !area[idx][c]) {
row[i][c] = col[j][c] = area[idx][c] = true;
} else {
return false;
}
}
}
return true;
}
}
9*9
的棋盘里,复杂度不随数据变化而变化。复杂度为9*9
的棋盘里,复杂度不随数据变化而变化。复杂度为这是我们「刷穿 LeetCode」系列文章的第 No.36
篇,系列开始于 2021/01/01,截止于起始日 LeetCode 上共有 1916 道题目,部分是有锁题,我们将先将所有不带锁的题目刷完。
在这个系列文章里面,除了讲解解题思路以外,还会尽可能给出最为简洁的代码。如果涉及通解还会相应的代码模板。
由于 LeetCode 的题目随着周赛 & 双周赛不断增加,为了方便我们统计进度,我们将按照系列起始时的总题数作为分母,完成的题目作为分子,进行进度计算。当前进度为 36/1916
。
为了方便各位同学能够电脑上进行调试和提交代码,我在 Github 建立了相关的仓库:https://github.com/SharingSource/LogicStack-LeetCode。在仓库地址里,你可以看到系列文章的题解链接、系列文章的相应代码、LeetCode 原题链接和一些其他的优选题解。