题目链接

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

题目原文

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.

题目翻译

判断一个数独是否是有效的。判断依据：Sudoku Puzzles – The Rules。
一个数独的表格可以是只填写了一部分，空的格子用点号’.’填充，比如，下图是一个有效的部分数独：

注意：一个有效的数独板（部分填充的）不一定可解。你只需要验证被填充的部分即可。

思路方法

首先，简单说一下规则：数独的每一行、每一列、9个九宫格里的数字只能是0-9，且不能重复。

思路一

最直观的思路，三个规则全部检查一遍即可。

代码

class Solution(object):
    def isValidSudoku(self, board):
        """ :type board: List[List[str]] :rtype: bool """
        for i in range(9):
            if not self.isValidNine(board[i]):
                return False
            col = [c[i] for c in board]
            if not self.isValidNine(col):
                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):
                    return False
        return True

    def isValidNine(self, row):
        map = {}
        for c in row:
            if c != '.':
                if c in map:
                    return False
                else:
                    map[c] = True
        return True

思路二

上面的代码虽然很好理解，但很冗余。我们可以用三个矩阵分别检查三个规则是否有重复数字，比如用row, col, block分别检查行、列、块是否有重复数字，代码如下：

代码

class Solution(object):
    def isValidSudoku(self, board):
        """ :type board: List[List[str]] :rtype: bool """
        # 注意：这里不能用[[False]*9]*9来初始化，牵涉到深浅拷贝的问题
        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
                    k = i/3*3 + j/3
                    if row[i][num] or col[j][num] or block[k][num]:
                        return False
                    row[i][num] = col[j][num] = block[k][num] = True
        return True

思路三

类似思路二，可以记录所有出现过的行、列、块的数字及相应位置，最后判断是否有重复。用set操作精简代码，可以有如下实现：

代码

class Solution(object):
    def isValidSudoku(self, board):
        """ :type board: List[List[str]] :rtype: bool """
        seen = []
        for i, row in enumerate(board):
            for j, c in enumerate(row):
                if c != '.':
                    seen += [(c,j),(i,c),(i/3,j/3,c)]
        return len(seen) == len(set(seen))

PS: 新手刷LeetCode，新手写博客，写错了或者写的不清楚还请帮忙指出，谢谢！
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