Leetcode 第36题：Valid Sudoku（有效的数独）

题目地址：Valid Sudoku

---

题目简介：

判断一个的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。

1. 数字1-9在每一行只能出现一次；
2. 数字1-9在每一列只能出现一次；
3. 数字1-9在每一个以粗实现分割的宫内只能出现一次；

                                                            

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。

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.

---

题目解析：寻找规律

参考网址：【LeetCode】36. Valid Sudoku

根据上述网址中的代码，得到了以下几个思路。

1. 首先，不考虑九宫格内是否有重复，需要考虑这个元素所在的行和列是否有重复的数字。这里需要遍历18个（当前位置看作两个数据）数据，所以可以用两个数组的位置来表示当前数字是否出现过。
2. 其次，考虑怎么进行遍历呢？如果每次遍历一列或者一行，那么需要遍历9行和9列。有一个想法是将矩阵对角线上的数据作为对称中心，第i行j列的数据对角线的数据为第j行i列，具体的想法如图：

剩下的有关九宫格的参见上面的链接；

C++代码：

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        for (int i = 0; i < 9; i++)
        {
            bool column[9], row[9], grid[9];
            memset(column,false,sizeof(column));
            memset(row,false,sizeof(row));
            memset(grid,false,sizeof(grid));
            for (int j = 0; j < 9; j++)
            {
                if (board[i][j] != '.')
                {
                    int temp = board[i][j] - '1';
                    if (row[temp])
                        return false;
                    row[temp] = true;
                }
                if (board[j][i] != '.')
                {
                    int temp = board[j][i] - '1';
                    if (column[temp])
                        return false;
                    column[temp] = true;
                }
                if(board[i/3*3+j/3][i%3*3+j%3] != '.')
                {
                    int temp = board[i/3*3+j/3][i%3*3+j%3] - '1';
                    if(grid[temp])
                        return false;
                    grid[temp] = true;
                }
            }
        }
        return true;
    }
};

---

Python代码：

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        for i in range(9):
            column = [False] * 9
            row = [False] * 9
            grid = [False] * 9
            
            for j in range(9):
                if board[i][j] != '.':
                    temp = int (board[i][j]) - int ('1')
                    if row[temp]:
                        return False
                    row[temp] = True;
                if board[j][i] != '.':
                    temp = int (board[j][i]) - int ('1')
                    if column[temp]:
                        return False
                    column[temp] = True
                if board[int(i/3)*3+int (j/3)][i%3*3+j%3] != '.':
                    temp = int (board[int(i/3)*3+int(j/3)][i%3*3+j%3]) - int ('1')
                    if(grid[temp]):
                        return False;
                    grid[temp] = True;
            
        return True