leetcode36题 题解 翻译 C语言版 Python版

36. Valid Sudoku

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

A partially filled sudoku which is valid.

Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

36.有效数独

判断一个数独是不是有效的。参考：数独规则

数独的边界能被部分填充，空的地方填充字符'.'。

如图是一个部分填充的数独并且是有效的。

注意：

一个有效的数独不一定有解。只需要验证填充的地方就行了。

思路：按规则来逐行逐列逐九宫格判断是不是有重复数字就行了。每次判断时新建一个数组，每找到一个数先判断数组中是否有这个数，有的话就无效，没有就填进去。因为数是0-9的，所以可以建一个长度为10的数组采用计数的方式让数字对应下标，也可以图省事用ascii码对应数组下标。

bool isValidSudoku(char** board, int boardRowSize, int boardColSize) {    for (int i = 0; i < boardRowSize; i++){        bool t[128] = {false};        for (int j = 0; j < boardColSize; j++){            if (board[i][j] == '.') continue;            if (t[board[i][j]]) return false;            else t[board[i][j]] = true;        }    }    for (int j = 0; j < boardColSize; j++){        bool t[128] = {false};        for (int i = 0; i < boardRowSize; i++){            if (board[i][j] == '.') continue;            if (t[board[i][j]]) return false;            else t[board[i][j]] = true;        }    }    for (int i = 0; i < 3; i++){        for (int j = 0; j < 3; j++){            bool t[128] = {false};            for (int m = 0; m < boardRowSize / 3; m++){                for (int n = 0; n < boardColSize / 3; n++){                    if (board[i*3+m][j*3+n] == '.') continue;                    if (t[board[i*3+m][j*3+n]]) return false;                    else  t[board[i*3+m][j*3+n]] = true;                }            }        }    }    return true;}

class Solution(object):    def isValidSudoku(self, board):        """        :type board: List[List[str]]        :rtype: bool        """        for line in board:            for i, cell in enumerate(line):                if cell != '.' and cell in line[:i]:                    return False        board2 = zip(*board)        for column in board2:            for i, cell in enumerate(column):                if cell != '.' and cell in column[:i]:                    return False        for i in range(3):            for j in range(3):                num = []                for m in range(3):                    for n in range(3):                        cell = board[i*3+m][j*3+n]                        if cell != '.':                            if cell in num:                                return False                            else: num.append(cell)        return True