[LeetCode] 36. 有效的数独
题目链接: https://leetcode-cn.com/problems/valid-sudoku/
题目描述:
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。
示例:
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:
首先还是理清题意,就是每一行,每一列,每一个小正方形都不能重复出现一个字母,如下图所示:
所以我们最直接想到就是,就是记录它的行,列,和小正方形的值,有重复就false
思路一:
我们用一个字典,分别记录行,列,和小正方形!
行,列我们直接可以用数字表示,小正方形如何表示呢?
这里,我们发现一个规律,我们可以把小正方形变成用二维数组唯一标识,比如(0,0)表示左上角那个,(1,1)表示中间那个,他们和行列的关系就是(i//3,j//3),
所以任何位置我们都能找出它在哪个行,哪个列,哪个正方形里!
时间复杂度都是常数级的
思路二:
上面我们用的空间复杂度有点多,要想办法改进空间复杂度,
我们有点小技巧,我们只需要用一个集合就可以搞定!
比如我们把board[i][j]
用字符串:
表示行:(i) + board[i][j]
表示列: board[i][j] + (j)
表示小正方形:(i) + board[i][j] + (j)
就直接可以用一个集合搞定!
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代码:
思路一
class Solution:
def isValidSudoku(self, board: List[List[str]]) -> bool:
from collections import defaultdict
row = defaultdict(set)
col = defaultdict(set)
small_square = defaultdict(set)
for i in range(9):
for j in range(9):
if board[i][j].isdigit():
if board[i][j] not in row[i] \
and board[i][j] not in col[j] \
and board[i][j] not in small_square[(i // 3,j // 3)]:
row[i].add(board[i][j])
col[j].add(board[i][j])
small_square[(i // 3, j // 3)].add(board[i][j])
else:
return False
return True
思路二
python
class Solution:
def isValidSudoku(self, board: List[List[str]]) -> bool:
one_set = set()
for i in range(9):
for j in range(9):
if board[i][j].isdigit():
row = "(" + str(i) + ")" + board[i][j]
col = board[i][j] + "(" + str(j) + ")"
small_square = "(" + str(i//3)+ ")" + board[i][j] + "(" + str(j//3) + ")"
if row in one_set or col in one_set or small_square in one_set:
return False
one_set.update([row,col,small_square])
return True
java
class Solution {
public boolean isValidSudoku(char[][] board) {
Set<String> one_set = new HashSet<>();
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
if (board[i][j] != '.') {
String row = "(" + i + ")" + board[i][j];
String col = board[i][j] + "(" + j + ")";
String small_square = "(" + i / 3 + ")" + board[i][j] + "(" + j / 3 + ")";
if (!one_set.add(row) || !one_set.add(col) || !one_set.add(small_square)) return false;
}
}
}
return true;
}
}
