36. 有效的数独 [media]

https://leetcode-cn.com/problems/valid-sudoku

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。

通过次数76,938 | 提交次数128,659

Second Try

参考题解用的小技巧，把matrix的编码写入一个循环中搞定，matrix的定位是一个小技巧，通过观察实现的。

rows, collumns，matrix都统一使用defaultdict的统计功能，其实相当于又把整个矩阵保存了三遍。

class Solution(object):
    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool

        关键是小矩阵里: i // 3 , j // 3相等的，就在同一个小矩阵里（0-2， 3-5，6-8竟然刚刚好，如果是1-9的反而不好处理，1-3，4-6，7-9）
        按行遍历，总共9个列表需要安全；按列遍历，也是9个列表需要安全

        关于矩阵如何定位有点麻烦，总共9个小矩阵，从0到9的位置排列刚好可以用自己的坐标来完成， row * 3 + columns, 刚刚好，属于观察得到的结论。
        matrix_idx = [(0, 0),(0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]

        """
        rows = [collections.defaultdict(int) for _ in range(9)]
        collumns = [collections.defaultdict(int) for _ in range(9)]
        matrix = [collections.defaultdict(int) for _ in range(9)]
        for r in range(9):
            for c in range(9):
                t = board[r][c]
                if t == ".":
                    continue
                rows[r][t] += 1
                collumns[c][t] += 1
                matrix[r // 3 * 3 + c // 3][t] += 1
                if any([rows[r][t] > 1, collumns[c][t] > 1, matrix[r // 3 * 3 + c // 3][t]> 1]):
                    return False     
        return True

执行用时：32 ms, 在所有 Python 提交中击败了98.54%的用户
内存消耗：12.7 MB, 在所有 Python 提交中击败了11.11%的用户

First Try

一开始不知道有什么高级内容，看了一会儿觉得还是直接用暴力破解，竟然也通过了，成绩还可以。不过就算有了思路，在写的时候竟然也是出现了一堆bug，调了一会儿，还是写得不够多。矩阵类的对某一行进行统计和对某一列进行统计难易程度是不同的，还好额外使用了空间把行列倒换来一波，事情就简单了许多。小矩阵的化，则是直接使用range进行跳转。

class Solution(object):
    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        def check_valid(matrix):
            clear_matrix = list(filter(lambda x: x!= ".", matrix))
            return len(clear_matrix) == len(set(clear_matrix))

        flag = True
        reverse_board = [[board[i][j] for i in range(9)] for j in range(9)]
        for i in range(9):
            if (not check_valid(board[i])) or (not check_valid(reverse_board[i])):
                flag = False
                break
        if not flag:
            return False
            
        for i in range(0, 7, 3):
            if not flag:
                break
            for j in range(0, 7, 3):
                matrix = [board[i + r][j + c] for r in range(0, 3) for c in range(0, 3)]
                if not check_valid(matrix):
                    flag = False
                    break
        return flag
            

执行用时 :52 ms, 在所有 Python 提交中击败了50.53%的用户
内存消耗 :12.7 MB, 在所有 Python 提交中击败了11.11%的用户

36. 有效的数独 [media]​

Second Try​

First Try​

36. 有效的数独 [media]

Second Try

First Try