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LeetCode: Valid Sudoku [035]

2024-07-05 04:31:37   230人阅读

【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.



【题意】

判断一个数独是否合法。一个合法的数独要求：每行、每列、9个小九宫格中都只能填充1~9九个数字各一次。


【思路】

1. 为每一行建一个行访问标记数组
2. 为每一列建一个列访问标记数组
3. 为每个九宫格建一个访问标记数组
扫描数独矩阵，凡是有数字填充的在相应的访问数组中标记，如果发现有重复标记的，则返回false


【代码】

class Solution {
public:
    bool isValidSudoku(vector<vector<char> > &board) {
        vector<vector<bool> > rows(9, vector<bool>(9, false));
        vector<vector<bool> > cols(9, vector<bool>(9, false));
        vector<vector<bool> > blocks(9, vector<bool>(9, false));
        for(int i=0; i<9; i++){
            for(int j=0; j<9; j++){
                if(board[i][j]==‘.‘)continue;
                //判断行
                if(rows[i][board[i][j]-‘0‘-1])return false;
                else rows[i][board[i][j]-‘0‘-1]=true;
                //判断列
                if(cols[j][board[i][j]-‘0‘-1])return false;
                else cols[j][board[i][j]-‘0‘-1]=true;
                //判断block
                if(blocks[i/3*3+j/3][board[i][j]-‘0‘-1])return false;       //注意block索引号的计算
                else blocks[i/3*3+j/3][board[i][j]-‘0‘-1]=true;
            }
        }
        return true;
    }
};


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