[开发工具][LeetCode] Valid Sudoku

Description

Valid Sudoku: Determine if a 9 x 9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

1. Each row must contain the digits 1-9 without repetition.
2. Each column must contain the digits 1-9 without repetition.
3. Each of the nine 3 x 3 sub-boxes of the grid must contain the digits 1-9 without repetition.

Example:

Input: 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"]]
Output: true

Solution

The basic idea is to verify if each row, each column, and each sub-box is valid.

We use the bits of a binary number to record the information of rows, columns, and sub-boxes. If a digits is appeared, we the the corresponding bit to 1. However, if a bit is already 1, we will kown that this row \ column \ sub-box is invalid.

So, the only thing is how to define different rows, columns, and sub-boxes. I use a diagram to illustrate my idea.

Code

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        vector<int> row(9, 0), col(9, 0), sub(9, 0);
        
        for(int i = 0; i < 9; i++) {
            for(int j = 0; j < 9; j++) {
                if(board[i][j] == '.')
                    continue;
                
                int num = board[i][j] - '0';
                if((row[i] & (1 << num)) || (col[j] & (1 << num)) || (sub[i/3 * 3 + j / 3] & (1 << num)))
                    return false;
                
                row[i] |= (1 << num);
                col[j] |= (1 << num);
                sub[i/3 * 3 + j / 3] |= (1 << num);
            }
        }
        return true;
    }
};

Complexity

Time complexity: O(1)

Space complexity: O(1)