LeetCode 36. 有效的数独（中）

首先，让我们来讨论下面两个关键问题，想清楚以下两个问题，这道题就很好理解并解决了：
1、如何枚举子数独？
可以使用 box_index = (row / 3) * 3 + columns / 3 ，其中 / 是整数除法。 2、如何确保行 / 列 / 子数独中没有重复项？
可以利用 value -> count 哈希映射来跟踪所有已经遇到的值。

题目

判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。
数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。

250px-Sudoku-by-L2G-20050714.svg.png

上图是一个部分填充的有效的数独。
数独部分空格内已填入了数字，空白格用 '.' 表示。

示例

示例 1: 

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输入:
[
  ["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"]
]
输出: true
示例 2: 
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输入:
[
  ["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"]
]
输出: false
解释: 除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。

考察知识点

哈希表

哪些情况下，可以用hash map？需要记录的情况很多、很复杂的时候，可以考虑使用hash map解决问题。

核心思想

方法一、一次迭代

首先，让我们来讨论下面两个关键问题，想清楚以下两个问题，这道题就很好理解并解决了：
1、如何枚举子数独？
可以使用 box_index = (row / 3) * 3 + columns / 3 ，其中 / 是整数除法。
5.png

2、如何确保行 / 列 / 子数独中没有重复项？
可以利用 value -> count 哈希映射来跟踪所有已经遇到的值。
36_slide_16.png

现在，我们完成了这个算法的所有准备工作：
- 遍历数独。 - 检查看到每个单元格值是否已经在当前的行 / 列 / 子数独中出现过： - 如果出现重复，返回 false。 - 如果没有，则保留此值以进行进一步跟踪。 - 返回 true。

代码步骤：

1、利用range迭代器初始化hash map：row、column、boxes，判断一个数独是否合法，要从这三个方面判断。数字 1-9 在每一个以粗实线分隔的 3x3 宫内、每一行、每一列只能出现一次。保存情况如下：row是一个list类型数据，rows[0]代表第一行情况，rows[0]是一个dict类型数据，rows[0] = {5: 1}代表5在第一行出现了1次。row、boxes同理。
2、遍历数独中的每一格，i是行号，j是列号。获取当前格子所在的 3x3 宫。
2.1、添加记录到row、column、boxes中
2.2、检查在行、列、3x3 小格子内，任何一种情况下出现次数超过1，就不是有效数独，返回False。
3、完成遍历没有问题则返回True。

LeetCode题解

Python版本

方法一实现

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from typing import List
class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        # init data 利用range迭代器初始化hashmap：row、column、boxes，判断一个数独是否合法，要从这三个方面判断。数字 1-9 在每一个以粗实线分隔的 3x3 宫内、每一行、每一列只能出现一次。保存情况如下：row是一个list类型数据，rows[0]代表第一行情况，rows[0]是一个dict类型数据，rows[0] = {5: 1}代表5在第一行出现了1次。row、boxes同理。
        rows = [{} for i in range(9)]
        columns = [{} for i in range(9)]
        boxes = [{} for i in range(9)]

        # validate a board  遍历数独中的每一格
        for i in range(9):  # i是行号 j是列号
            for j in range(9):
                num = board[i][j]
                if num != '.':
                    num = int(num)
                    box_index = (i // 3 ) * 3 + j // 3  # 获取当前格子所在的 3x3 宫
                    
                    # keep the current cell value
                    rows[i][num] = rows[i].get(num, 0) + 1  # 行出现次数加一
                    columns[j][num] = columns[j].get(num, 0) + 1  # 列出现次数加一
                    boxes[box_index][num] = boxes[box_index].get(num, 0) + 1  # 3x3 宫内出现次数加一
                    
                    # check if this value has been already seen before
                    if rows[i][num] > 1 or columns[j][num] > 1 or boxes[box_index][num] > 1:  # 如果在行、列、3x3 小格子内，任何一种情况下出现次数超过1，就不是有效数独，返回False
                        return False         
        return True  # 完成遍历没有问题 返回True


print("leet code accept!!!")
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"]
    ],
    [
        ["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"]
    ]
]
Answer = [True, False]

if __name__ == "__main__":
    solution = Solution()
    for i in range(len(Input)):
        print("-"*50)
        result = solution.isValidSudoku(Input[i])
        print(result==Answer[i])

时间复杂度：由于只对81个单元格进行遍历，所以时间复杂度为\(O(1)\)。
空间复杂度： \(O(1)\)。
执行用时 :44 ms, 在所有 Python3 提交中击败了98.28%的用户。
内存消耗 :13.4 MB, 在所有 Python3 提交中击败了34.03%的用户。