力扣 36.有效的数独

36. 有效的数独

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。

官方

链接

方法一：一次遍历
有效的数独满足以下三个条件：

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        int rows[9][9];
        int columns[9][9];
        int subboxes[3][3][9];
        
        memset(rows,0,sizeof(rows));
        memset(columns,0,sizeof(columns));
        memset(subboxes,0,sizeof(subboxes));
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                char c = board[i][j];
                if (c != '.') {
                    int index = c - '0' - 1;
                    rows[i][index]++;
                    columns[j][index]++;
                    subboxes[i / 3][j / 3][index]++;
                    if (rows[i][index] > 1 || columns[j][index] > 1 || subboxes[i / 3][j / 3][index] > 1) {
                        return false;
                    }
                }
            }
        }
        return true;
    }
};

大神思路

链接

与官方题解不一样在

遍历到每个数的时候，例如boar[i][j]，我们判断其是否满足三个条件：

在第 i 个行中是否出现过
在第 j 个列中是否出现过
在第 j/3 + (i/3)*3个box中是否出现过.为什么是j/3 + (i/3)*3呢？
关于从数组下标到box序号的变换
 重述一遍问题：给定i和j，如何判定board[i][j]在第几个box呢？
 显然属于第几个box由i和j的组合唯一确定，例如board[2][2]一定是第0个box，board[4][7]一定是第5个box，可以画出来看一下，但是规律在哪里呢？
我们可以考虑一种简单的情况： 一个3x9的矩阵，被分成3个3x3的box，如图：

显然每个数属于哪个box就只取决于纵坐标，纵坐标为0/1/2的都属于box[0],纵坐标为3/4/5的都属于box[1],纵坐标为6/7/8的都属于box[2].也就是j/3.
而对于9x9的矩阵，我们光根据j/3得到0/1/2还是不够的，可能加上一个3的倍数，例如加0x3,表示本行的box，加1x3，表示在下一行的box，加2x3，表示在下两行的box， 这里的0/1/2怎么来的？和j/3差不多同理，也就是i/3。

合并

利用大神处理subbox的思路 j/3 + (i/3)*3

优化一下subboxes[3][3][9]

 变成二维

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        int rows[9][9];
        int columns[9][9];
        int subboxes[9][9];
        
        memset(rows,0,sizeof(rows));
        memset(columns,0,sizeof(columns));
        memset(subboxes,0,sizeof(subboxes));
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                char c = board[i][j];
                if (c != '.') {
                    int index = c - '0' - 1;
                    rows[i][index]++;
                    columns[j][index]++;
                    subboxes[j/3 + (i/3)*3][index]++;
                    if (rows[i][index] > 1 || columns[j][index] > 1 || subboxes[j/3 + (i/3)*3][index] > 1) {
                        return false;
                    }
                }
            }
        }
        return true;
    }
};