LeetCode数组-有效的数独

最新推荐文章于 2023-02-07 16:53:45 发布

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最新推荐文章于 2023-02-07 16:53:45 发布

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分类专栏： LeetCode

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本文介绍了一个高效的算法，用于判断9x9数独的有效性。通过检查数字1-9在每行、每列以及每个3x3宫内的唯一性，确保数独符合基本规则。文章提供了详细的实现代码和示例。

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有效的数独
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。

?我不会

大神的代码：

        dic_row = [{},{},{},{},{},{},{},{},{}]
        dic_col = [{},{},{},{},{},{},{},{},{}]
        dic_box = [{},{},{},{},{},{},{},{},{}]

# 一次遍历，根据 i、j可知道每个值的位置 属于哪个3 x 3 , 行、列
        for i in range(len(board)):
            for j in  range(len(board)):
                num = board[i][j]
                if num == ".":
                    continue
                elif 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 True