leetcode36. Valid Sudoku

最新推荐文章于 2022-07-31 21:20:38 发布

原创最新推荐文章于 2022-07-31 21:20:38 发布 · 363 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 同时被 2 个专栏收录

32 篇文章

订阅专栏

数组

12 篇文章

订阅专栏

本文介绍了一种简单有效的算法来验证数独是否有效。通过检查数独的每一行、每一列以及每一个九宫格内的数字是否重复，可以快速判断一个数独是否符合基本规则。

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.

比较简单，只需要检查细节就行了

AC解：

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) 
    {
        bool used[10];
       
        //检查行
        for (int i = 0; i < 9; i++)
        {
            fill(used,used + 10,false);
            for (int j = 0; j < 9; j++)
            {
                if (!check(used,board[i][j]))
                    return false;
            }
        }
        //检查列
        for (int i = 0; i < 9; i++)
        {
            fill(used,used + 10,false);
            for (int j = 0; j < 9; j++)
                if (!check(used,board[j][i]))
                    return false;
        }
        //检查九方格
        for (int i = 0; i < 3; i++)
            for (int j = 0; j < 3; j++)
            {
                fill(used,used + 10,false);
                for (int m = 3 * i; m < 3 * i + 3; m++)
                    for (int n = 3 * j; n < 3 * j + 3; n++)
                        if (!check(used,board[m][n]))
                            return false;
            }
        
        return true;
    }
    
    bool check(bool used [],char a)
    {
        if (a == '.')    return true;
        if (used[a - '0']) return false;
        
        used[a - '0'] = true;
        return true;
    }
};