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。
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。
- 数字
我的解法——遍历(AC)
- 遍历一遍
- 时间空间复杂度均为O(n)。
- Beat41%
class Solution:
def isValidSudoku(self, board):
"""
:type board: List[List[str]]
:rtype: bool
"""
for i in range(9):
for j in range(9):
if(board[i][j] == "."):
board[i][j] = 0
else:
board[i][j] = int(board[i][j])
for i in range(9):
now = [1] * 10
for j in range(9):
if(board[i][j] != 0):
if(now[board[i][j]] <= 0):
return False
else:
now[board[i][j]] -= 1
for j in range(9):
now = [1] * 10
for i in range(9):
if(board[i][j] != 0):
if(now[board[i][j]] <= 0):
return False
else:
now[board[i][j]] -= 1
for i in range(3):
for j in range(3):
now = [1] * 10
for m in range(3):
for n in range(3):
if(board[i * 3 + m][j * 3 + n] != 0):
if(now[board[i * 3 + m][j * 3 + n]] <= 0):
return False
else:
now[board[i * 3 + m][j * 3 + n]] -= 1
return True
最佳解法——遍历+Hash
- HashSet——利用i和j,分别构造行列和块。i代表行,j代表列,即可构造每行的情况;i代表列,j代表行,即可构造每列的情况;利用i构造每块的首个元素编号,代表每个块,再利用j构造块中的每个元素。很机智的做法,就不需要设置很多HashSet。
- 只遍历一遍,我的方法遍历了三遍!
class Solution:
def isValidSudoku(self, board):
"""
:type board: List[List[str]]
:rtype: bool
"""
dic_row = [{},{},{},{},{},{},{},{},{}]
dic_col = [{},{},{},{},{},{},{},{},{}]
dic_box = [{},{},{},{},{},{},{},{},{}]
for i in range(len(board)):
for j in range(len(board)):
num = board[i][j]
if num == ".":
continue
if num not in dic_row[i] and num not in dic_col[j] and num not in dic_box[3*(i//3)+(j//3)]:
dic_row[i][num] = 1
dic_col[j][num] = 1
dic_box[3*(i//3)+(j//3)][num] = 1
else:
return False
return Truepublic 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;
}
参考答案
- https://segmentfault.com/a/1190000009297312