[LeetCode] Valid Sudoku

最新推荐文章于 2021-05-09 13:31:40 发布

原创最新推荐文章于 2021-05-09 13:31:40 发布 · 315 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

131 篇文章

订阅专栏

本文介绍了一种通过检查行、列及每个3x3宫格内的数字是否重复来验证数独盘面有效性的算法实现。该算法适用于部分填充的数独盘面，且不考虑数独的可解性。

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.

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

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