LeetCode:36.有效的数独

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36.有效的数独

来源：力扣（LeetCode）

链接: leetcode.cn/problems/va…

请你判断一个 9 x 9 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。（请参考示例图）

注意：

• 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。
• 空白格用 '.' 表示。

示例 1：

输入：board = 
[["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：board = 
[["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。

提示：

• board.length == 9
• board[i].length == 9
• board[i][j] 是一位数字（1-9）或者 '.'

解法

• 循环遍历即可，为了避免同一行同一列或者同一个九宫格中有相同的元素，可以采用三个set列表，表示每行，每列，每个九宫格中出现的元素，遍历到新的元素的时候，先看看是否在这些对应的集合中出现过，如果出现过，返回False， 如果没有出现过，将该元素加入到对应的集合中，并继续遍历，最终返回True。
• 这里注意：当遍历到某点(i,j)时，辨别其是第几个九宫格的计算公式为： $index = i / 3 * 3 + j / 3$

代码实现

遍历+剪枝

python实现

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        self.rows = [set() for _ in range(9)]
        self.cols = [set() for _ in range(9)]
        self.boards = [set() for _ in range(9)]

        for i in range(9):
            for j in range(9):
                ele = board[i][j]
                if ele == '.':
                    continue
                index = i // 3 * 3 + j // 3
                if ele in self.rows[i] or ele in self.cols[j] or ele in self.boards[index]: 
                    return False

                self.rows[i].add(ele)
                self.cols[j].add(ele)
                self.boards[index].add(ele)
        return True

c++实现

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        vector<set<char>> rows(9); 
        vector<set<char>> cols(9); 
        vector<set<char>> boards(9); 

        int m = board.size(), n = board[0].size();
        for(int i=0; i<m; i++) {
            for(int j=0; j<n; j++) {
                char ele = board[i][j];
                int index = i/3*3 + j/3;
                if (ele == '.') continue;

                if (rows[i].count(ele) != 0 || cols[j].count(ele) != 0 || boards[index].count(ele) != 0) return false;
                rows[i].insert(ele);
                cols[j].insert(ele);
                boards[index].insert(ele);
            }
        }
        return true;
    }
};

复杂度分析

• 时间复杂度： $O(mn)$
• 空间复杂度： $O(mn)$

参考

• leetcode.cn/problems/va…