36. Valid Sudoku

Leetcode Hash Table

题目

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

• Each row must contain the digits 1-9 without repetition.
• Each column must contain the digits 1-9 without repetition.
• 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.

思路

利用hash table(这里是字典)，识别是否出现数字重复。利用numpy方便处理矩阵元素。

class Solution(object):

    def no_repetation(self, alist):
        n = len(alist)
        mydict = {"1":0,"2":0,"3":0, "4":0,"5":0, "6":0, "7":0, "8":0, "9":0}

        for x in alist:
            if x == '.':
                continue
            if mydict[x] == 0:
                    mydict[x] = 1
            else:
                return False

        return True


    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        import numpy as np
        import copy

        myboard = np.array(board)

        for i in range(9):

            # Each row, column must contain the digits 1-9 without repetition.
            eachrow = myboard[i,:]
            repetation = self.no_repetation(eachrow)
            if not repetation:
                return False

            # Each row, column must contain the digits 1-9 without repetition.
            eachcolumn = myboard[:,i]
            repetation = self.no_repetation(eachcolumn)
            if not repetation:
                return False

        # Each of the 9 3x3 sub-boxes of the grid must contain the digits 1-9 without repetition.
        for i in range(3):
            for j in range(3):
                eachblock = myboard[i*3:(i+1)*3, j*3:(j+1)*3].reshape((-1,))
                repetation = self.no_repetation(eachblock)
                if not repetation:
                    return False

        return True

还有一种更巧妙的方法，就是每一行、每一列、每一块都是一个字典预先放在列表中；并且对于块的处理，巧妙的利用矩阵元素的行列数计算所在的块：

class Solution:
    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        dic_row = [{},{},{},{},{},{},{},{},{}]
        dic_col = [{},{},{},{},{},{},{},{},{}]
        dic_box = [{},{},{},{},{},{},{},{},{}]

        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
                else:
                    return False

        return True

对应的C++版本为：

bool isValidSudoku(vector<vector<char> > &board)
    {
        int used1[9][9] = {0}, used2[9][9] = {0}, used3[9][9] = {0};

        for(int i = 0; i < board.size(); ++ i)
            for(int j = 0; j < board[i].size(); ++ j)
                if(board[i][j] != '.')
                {
                    int num = board[i][j] - '0' - 1, k = i / 3 * 3 + j / 3;
                    if(used1[i][num] || used2[j][num] || used3[k][num])
                        return false;
                    used1[i][num] = used2[j][num] = used3[k][num] = 1;
                }

        return true;
    }