题目

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

A partially filled sudoku which is valid.

Note:

A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

翻译

数独嘛 略

Hints

Related Topics: Hash Table

通过哈希表实现O(n^2)的算法 即只遍历整个数独的表一遍就可以得到有效性判断

代码

C++

class Solution

{

public:

    bool isValidSudoku(vector<vector<char> > &board)

    {

        int used1[9][9] = {0}, used2[9][9] = {0}, used3[9][9] = {0};

        for(int i = 0; i < board.size(); ++ i)

            for(int j = 0; j < board[i].size(); ++ j)

                if(board[i][j] != '.')

                {

                    int num = board[i][j] - '0' - 1, k = i / 3 * 3 + j / 3;

                    if(used1[i][num] || used2[j][num] || used3[k][num])

                        return false;

                    used1[i][num] = used2[j][num] = used3[k][num] = 1;

                }

        return true;

    }

};

Java

//an interesting solution from discuss

public boolean isValidSudoku(char[][] board) {

    Set seen = new HashSet();

    for (int i=0; i<9; ++i) {

        for (int j=0; j<9; ++j) {

            char number = board[i][j];

            if (number != '.')

                if (!seen.add(number + " in row " + i) ||

                    !seen.add(number + " in column " + j) ||

                    !seen.add(number + " in block " + i/3 + "-" + j/3))

                    return false;

        }

    }

    return true;

}

Python

class Solution(object):

    def isValidSudoku(self, board):

        """

        :type board: List[List[str]]

        :rtype: bool

        """

        row = {}

        col = {}

        cube = {}

        for i in range(0,9):

            row[i] = set()

            for j in range(0,9):

                if j not in col:    col[j] = set()

                if board[i][j] !='.':

                    num = int(board[i][j])

                    k = i/3*3+j/3

                    if k not in cube:   cube[k] = set()

                    if num in row[i] or num in col[j] or num in cube[k]:

                        return False

                    row[i].add(num)

                    col[j].add(num)

                    cube[k].add(num)

        return True