Leetcode 036 有效的数独 思路详解 python

本人一直在努力地积累Leetcode上用Python实现的题，并且会尽力讲清每道题的原理，绝不像其他某些博客简略地带过。

如果觉得讲的清楚，欢迎关注。

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。

1. 数字 1-9 在每一行只能出现一次。
2. 数字 1-9 在每一列只能出现一次。
3. 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。

思路还是比较清晰，因为这道题没有什么其他的解法。主要方法可以总结为：1.检查每一行是否符合规范。

2.检查每一列

3.检查每一个3*3矩阵

接下来贴一下本人的代码，以及一位大神的代码：

本人：

class Solution:
    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        #每个3*3矩阵中心在二维数组中的坐标
        self.center = [(1,1), (1,4), (1,7), (4,1), (4,4), (4,7), (7,1), (7,4), (7,7)]
        #先假定这是有效的
        ans = True
        #用一个函数检查行和列
        for i in range(9):
            ans = self.judgelineandrow(board, i)
            if ans is False:
                break

        if ans is False:
            return ans

        ans = self.judgesqure(board)
        return ans

        


    def judgelineandrow(self, board, i):
        #通过数数字与最终集合长度的对比，判断有没有一样的
        rowcount = 0
        colcount = 0
        rowset = set()
        colset = set()
        ans = True
        for j in range(9):
            if board[i][j] != '.':
                colcount += 1
                colset.add(board[i][j])
            if board[j][i] != '.':
                rowcount += 1
                rowset.add(board[j][i])
        if len(rowset) != rowcount or len(colset) != colcount:
            ans = False
        return ans

    def judgesqure(self, board):
        ans = True
        #提取中心点坐标
        for i , j in self.center:
            squareset = set()
            numbercount = 0
            for k in range(-1, 2):
                for n in range(-1, 2):
                    #利用几何性质不断检查其他坐标的值
                    if board[i-k][j-n] != '.':
                        numbercount += 1
                        squareset.add(board[i-k][j-n])
            if len(squareset) != numbercount:
                ans = False
                return ans
        return ans

我的优点在于分了函数，结构还是比较清晰的。但是检查的逻辑有点傻，其他人写的基本是通过一个字符串，遍历时不断增加这个

字符串，然后检查当前的字符在不在字符串内。如果在就无效。

大神：

class Solution:
    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        #每一个字典表示一行的元素出现次数

        dic_row = [{},{},{},{},{},{},{},{},{}]            # 每行的元素以一个字典储存，key是数字，value统一为1.
        dic_col = [{},{},{},{},{},{},{},{},{}]
        dic_box = [{},{},{},{},{},{},{},{},{}]


        #只遍历一遍整个棋盘
        #这里比较强大的地方在于它可以通过i,j来推断这个点在哪个3*3网格内
        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       # 利用地板除，向下取余。巧妙地将矩阵划分为九块
                else:
                     return False

        return True

非常优美，respect。