leetcode36. Valid Sudoku

验证数独的有效性

最新推荐文章于 2022-07-31 21:20:38 发布

weixin_30404405

最新推荐文章于 2022-07-31 21:20:38 发布

阅读量37

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/wxquare/p/6030690.html

本文介绍了一种验证数独是否有效的算法实现。该算法检查数独盘面的每一行、每一列以及每一个宫格内数字1到9是否只出现一次。即使数独未完全填满也可以进行验证。

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.

class Solution {
public:
    bool isValidSudoku(vector<vector<char>> & board) {
        int N = board.size();
        vector<vector<bool>> used1(N,vector<bool>(N,false));
        vector<vector<bool>> used2(N,vector<bool>(N,false));
        vector<vector<bool>> used3(N,vector<bool>(N,false));

        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                if(board[i][j] != '.'){
                    int num = board[i][j] - '0' - 1;
                    int k = i/3*3+j/3;
                    if(used1[num][j] || used2[i][num] || used3[k][num])
                        return false;
                    else
                        used1[num][j] = true;
                        used2[i][num] = true;
                        used3[k][num] = true;
                }
            }
        }
        return true;
    }
};