【LeetCode】36. Valid Sudoku 解题报告（Python）

最新推荐文章于 2020-07-17 17:40:41 发布

翻译最新推荐文章于 2020-07-17 17:40:41 发布 · 444 阅读

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本文介绍了一种使用Python实现的数独有效性验证算法。通过遍历数独的每一行、每一列以及每一个小格子，利用字典来检查数字是否重复，从而判断数独是否有效。文章详细解释了如何计算小格子的坐标，并提供了完整的代码示例。

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题目分析：

这一题比较简单，甚至我觉得有耐心可以不用循环写出来（滑稽），用循环就时取出每行、每列、每个小格子里非’.'的值，再判断是否有重复值即可，可以用set或者dict取判断，我觉得dict要简单一些，不存在就加进去，存在返回False。
在这里插入图片描述
关于行列没什么说的，而小格子的坐标有点难度，[int(i/3)*3 + int(j/3)]可以理解为第几个小格子，[int(i%3)*3 + int(j%3)]可以理解为小格子中的第几个元素，你可以写几个格子的坐标找一下这个规律。

测试代码：

class Solution:
    def isValidSudoku(self, board):
        for i in range(9):
            #分别储存一行，一列，一个小方格
            heng, shu, fang = {}, {}, {}
            for j in range(9):
                #行,首先排除'.'
                if board[i][j] != '.':
                    #不在则加入字典，在了就是重复返回False
                    if board[i][j] not in heng: heng.update({board[i][j]:0})
                    else:   return False
                #列,首先排除'.'
                if board[j][i] != '.':
                    # 不在则加入字典，在了就是重复返回False
                    if board[j][i] not in shu: shu.update({board[j][i]:0})
                    else:   return False
                #小方格,首先排除'.'，[int(i/3)*3 + int(j/3)]可以理解为第几个小格子，[int(i%3)*3 + int(j%3)]可以理解为小格子中的第几个元素
                if board[int(i/3)*3 + int(j/3)][int(i%3)*3 + int(j%3)] != '.':
                    # 不在则加入字典，在了就是重复返回False
                    if board[int(i/3)*3 + int(j/3)][int(i%3)*3 + int(j%3)] not in fang: fang.update({board[int(i/3)*3 + int(j/3)][int(i%3)*3 + int(j%3)]:0})
                    else:   return False
        return True

print(Solution().isValidSudoku([
  ["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"]
])) #提交时请删除print