36、有效的数独

1）题目描述

请你判断一个 9 x 9 的数独是否有效。只需要 根据以下规则 ，验证已经填入的数字是否有效即可。

1. 数字 1-9 在每一行只能出现一次。
2. 数字 1-9 在每一列只能出现一次。
3. 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。（请参考示例图）

注意：

• 一个有效的数独（部分已被填充）不一定是可解的。
• 只需要根据以上规则，验证已经填入的数字是否有效即可。
• 空白格用 '.' 表示。

示例 1：

输入：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"]]
输出：true

示例 2：

输入：board = 
[["8","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"]]
输出：false
解释：除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。 但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。

提示：

• board.length == 9
• board[i].length == 9
• board[i][j] 是一位数字（1-9）或者 '.'

2）分析

哈希表一次遍历。可以使用哈希表记录每一行、每一列和每一个小九宫格中，每个数字出现的次数。只需要遍历数独一次，在遍历的过程中更新哈希表中的计数，并判断是否满足有效的数独的条件即可。

对于给i行j列的元素，其行位置为i，列位置为j，小方格位置i/3,j/3。设定三个数组分别r[9][9]、l[9][9]、box[3][3][9]记录行、列、小方格中1-9九个数字出现的次数。r和l都是二位是数组，有九行，每一行九个元素，分别记录数独表中每一行里面1-9的出现次数和数独表中每一列里面1-9的出现次数，box是三维数组，前两维表示小方格在数独表中的坐标（三行三列），第三维记录每个小方格中1-9出现的次数。

遍历过程中，board[i][j]–>r[i][value]++,l[j][value]++,box[i/3][j/3][value]++，其中value是board[i][j]对应的数字。同时判断自增之后的r[i][value]++,l[j][value]++,box[i/3][j/3][value]++若大于一，则返回false。若遍历结束都没有出现大于一，返回true。

3）C++代码

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        int r[9][9];
        int l[9][9];
        int box[3][3][9];

        memset(r,0,sizeof(r));
        memset(l,0,sizeof(l));
        memset(box,0,sizeof(box));

        for(int i=0;i<9;i++){
            for(int j=0;j<9;j++){
                char ch=board[i][j];
                if(ch!='.'){
                    int value=ch-'0'-1;
                    r[i][value]++;
                    l[j][value]++;
                    box[i/3][j/3][value]++;
                    if(r[i][value]>1||l[j][value]>1||box[i/3][j/3][value]>1)
                        return false;
                }
            }
        }
        return true;
    }
};