LeetCode算法题36：有效的数独解析

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。

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 1:

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:

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。

说明:

一个有效的数独（部分已被填充）不一定是可解的。
只需要根据以上规则，验证已经填入的数字是否有效即可。
给定数独序列只包含数字 1-9 和字符 ‘.’ 。
给定数独永远是 9x9 形式的。

这个题就按标记查看行、列、3*3单元格是否数字重复即可，遍历整个表时遇到数字将对应标志数组的位置置为true，如果下次发现位置已经为true说明重复，返回false。行列的标志很好置位，单元格的数组置位可以这样：一共九个单元格，其坐标与数组对应位置为

(0，0)(0，1)(0，2)			(0，3)(0，4)(0，5)			(0，6)(0，7)(0，8)
(1，0)(1，1)(1，2)-->第0行	(1，3)(1，4)(1，5)-->第1行	(1，6)(1，7)(1，8)-->第2行
(2，0)(2，1)(2，2)			(2，3)(2，4)(2，5)			(2，6)(2，7)(2，8)
(3，0)(3，1)(3，2)			(3，3)(3，4)(3，5)			(3，6)(3，7)(3，8)
(4，0)(4，1)(4，2)-->第3行	(4，3)(4，4)(4，5)-->第4行	(4，6)(4，7)(4，8)-->第5行
(5，0)(5，1)(5，2)			(5，3)(5，4)(5，5)			(5，6)(5，7)(5，8)
(6，0)(6，1)(6，2)			(6，3)(6，4)(6，5)			(6，6)(6，7)(6，8)
(7，0)(7，1)(7，2)-->第8行	(7，3)(7，4)(7，5)-->第7行	(7，6)(7，7)(7，8)-->第8行
(8，0)(8，1)(8，2)			(8，3)(8，4)(8，5)			(8，6)(8，7)(8，8)

所以其对应关系可以描述为3*(i/3)+(j/3)。

C++源代码：

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        int m = board.size();
        int n = board[0].size();
        if (m!=9 || n!=9) return false;
        vector<vector<bool>> row(m, vector<bool>(n, false));
        vector<vector<bool>> col(m, vector<bool>(n, false));
        vector<vector<bool>> cell(m, vector<bool>(n, false));
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                int t = board[i][j];
                if(t>='1' && t<='9')
                {
                    int k = t - '1';
                    if(row[i][k] || col[k][j] || cell[3*(i/3)+j/3][k])
                        return false;
                    row[i][k] = true;
                    col[k][j] = true;
                    cell[3*(i/3)+j/3][k] = true;
                }
            }
        }
        return true;
    }
};

python3源代码：

class Solution:
    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        m = len(board)
        n = len(board[0])
        if m!=9 or n!=9:
            return False
        row = [[False for col in range(n)] for row in range(m)]
        col = [[False for col in range(n)] for row in range(m)]
        cell = [[False for col in range(n)] for row in range(m)]
        for i in range(m):
            for j in range(n):
                t = ord(board[i][j])
                if t>=ord('1') and t<=ord('9'):
                    k = t - ord('1')
                    if row[i][k] or col[k][j] or cell[3*(i//3)+(j//3)][k]:
                        return False
                    row[i][k] = True
                    col[k][j] = True
                    cell[3*(i//3)+(j//3)][k] = True
        return True

这里要注意的是python中list初始化的时候的代码，这种初始化是比较推荐的，如果使用[[false]*n]*m这种方式只是浅复制，会造成只要修改一个那么每一行都会修改的问题。所以还是采用推荐的方法去初始化列表。

LeetCode算法题36：有效的数独解析

猜你喜欢