python - 遍历有效数独的子框

Question

我正在研究Valid Sudoku - LeetCode并且无法弄清楚为什么 box_index = (i//3 ) * 3 + j//3 能够遍历子框

Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

1. Each row must contain the digits 1-9 without repetition.
2. Each column must contain the digits 1-9 without repetition.
3. Each of the 9 3x3 sub-boxes of the grid must contain the digits 1-9 without repetition.

A partially filled sudoku which is valid.

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

Example 1:

Input:
[
  ["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"]
]
Output: true

Example 2:

Input:
[
  ["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"]
]
Output: false
Explanation: Same as Example 1, except with the 5 in the top left corner being 
    modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.

Note:

• A Sudoku board (partially filled) could be valid but is not necessarily solvable.
• Only the filled cells need to be validated according to the mentioned rules.
• The given board contain only digits 1-9 and the character '.'.
• The given board size is always 9x9.

阅读一个聪明的解决方案

class Solution:
    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        # init data
        rows = [{} for i in range(9)]
        columns = [{} for i in range(9)]
        boxes = [{} for i in range(9)]

        # validate a board
        for i in range(9):
            for j in range(9):
                num = board[i][j]
                if num != '.':
                    num = int(num)
                    box_index = (i // 3 ) * 3 + j // 3

                    # keep the current cell value
                    rows[i][num] = rows[i].get(num, 0) + 1
                    columns[j][num] = columns[j].get(num, 0) + 1
                    boxes[box_index][num] = boxes[box_index].get(num, 0) + 1

                    # check if this value has been already seen before
                    if rows[i][num] > 1 or columns[j][num] > 1 or boxes[box_index][num] > 1:
                        return False         
        return True

测试用例

class MyCase(unittest.TestCase):
   class MyCase(unittest.TestCase):
    def setUp(self):
        self.solution = Solution()

    def test_a(self):
        board = [   ["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"]
                ]
        check = self.solution.isValidSudoku(board)
        self.assertTrue(check)

    def test_b(self):
        board = [   ["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"]
                ]
        check = self.solution.isValidSudoku(board)
        self.assertFalse(check)

unittest.main() def setUp(self):
        self.solution = Solution()

    def test_a(self):
        board = [   ["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"]
                ]
        check = self.solution.isValidSudoku(board)
        self.assertTrue(check)

    def test_b(self):
        board = [   ["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"]
                ]
        check = self.solution.isValidSudoku(board)
        self.assertFalse(check)

unittest.main()

能否请您提供一些提示，为什么 box_index = (i//3 ) * 3 + j//3 可以遍历子盒子？

Answer 1

您可以将整个盒子拆分为3*3子盒子，(i, j)属于索引为的子盒子>(i//3, j//3)，这是一个 3*3 二维数组。如果我们想将其展平为 1*9 一维数组，索引将为 (i//3 ) * 3 + j//3。

带有索引的子框:

|0|1|2|
|3|4|5|
|6|7|8|

如果你仍然感到困惑，你可以尝试一些例子，然后弄清楚。

希望对您有帮助，如有疑问请评论。 :)