LeetCode Medium: 36. Valid Sudoku

一、题目

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.

题目大致意思就是，数独的规则，每行1~9不重复，每列1~9不能重复，3x3小方格1~9不重复。

二、思路

此题使用可以分别判断行列和小方格内的数字是否重复，可以利用哈希思想，其实就是python中的字典也是一样的，逐一遍历行列和小方格，遇到不同的值就存进去，若是存储的过程中遇到相同的就返回False。

行号规律

观察行号规律：

第0个九宫格：000111222; 第1个九宫格：000111222; 第2个九宫格：000111222;

第3个九宫格：333444555; 第4个九宫格：333444555; 第5个九宫格：333444555;

第6个九宫格：666777888; 第7个九宫格：666777888; 第8个九宫格：666777888;

可见对于每三个九宫格行号增3；对于单个九宫格，每三个格点行号增1。

因此第i个九宫格的第j个格点的行号可表示为i/3*3+j/3

三、代码

#coding:utf-8
‘‘‘
Solution0最原始暴力求解，但代码冗余
‘‘‘
class Solution0:
    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        for row in range(9):
            if not self.isValidNine(board[row]):
                return False
            column = [c[row] for c in board]
            if not self.isValidNine(column):
                print(‘False‘)
                return False
        for i in [0,3,6]:
            for j in [0,3,6]:
                block = [board[s][t] for s in [i,i+1,i+2] for t in [j,j+1,j+2]]
                if not self.isValidNine(block):
                    print(False)
                    return False
        print(‘True‘)
        return True

    def isValidNine(self,row):
        """
        当前九个数字是否有效
        :param row:int
        :return:bool
        """
        map = {}
        for c in row:
            if c != ‘.‘:
                if c in map:
                    return False
                else:
                    map[c] = True #若字典map中没有c，则将c存至该map字典中，key为c，value为True
        return True

‘‘‘
Solution1精简代码，分别用三个矩阵来检查三个规则是否有重复数字
‘‘‘
class Solution1:
    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        row = [[False for i in range(9)] for j in range(9)]
        col = [[False for i in range(9)] for j in range(9)]
        block = [[False for i in range(9)] for j in range(9)]
        for i in range(9):
            for j in range(9):
                if board[i][j] != ‘.‘:
                    num = int(board[i][j])-1  #这里其实使用了哈希思想，因为数字都是在0~9之间，所以索引减1，试想如果某个位置上的数字和其他位置的数字相同的话，其num值也是一样的，所以通过True或者False就可以辨别是否有重复
                    k = i//3*3 + j//3    #K表示第i个九宫格的第j个格点的行号
                    if row[i][num] or col[j][num] or block[k][num]:
                        print(‘False‘)
                        return False
                    row[i][num] = col[j][num] = block[k][num] = True
        print(‘True‘)
        return True

if __name__ == ‘__main__‘:
    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"]
]
    #ss =Solution0()
    ss = Solution1()
    ss.isValidSudoku(board)

　参考博客：https://www.cnblogs.com/ganganloveu/p/4170632.html      https://blog.csdn.net/coder_orz/article/details/51596499

原文地址：https://www.cnblogs.com/xiaodongsuibi/p/8982486.html