LeetCode Medium: 36. Valid Sudoku
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一、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 digits1-9
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
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