本文已参与「新人创作礼」活动，一起开启掘金创作之路。
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。
- 数字 1-9 在每一行只能出现一次。
- 数字 1-9 在每一列只能出现一次。
- 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 '.'
解题思路
思路就是每个位置只遍历一次，无论横着切、竖着切、块着切，都是讲将总体分为9部分，每部分不能有重复的。
因此每部分单独有一个9位数组来存是否重复
代码实现
class Solution {
public:
bool isValidSudoku(vector<vector<char>>& board) {
int c_arr[9][9];
int r_arr[9][9];
int m_arr[3][3][9];
memset(c_arr,0,sizeof(c_arr));
memset(r_arr,0,sizeof(r_arr));
memset(m_arr,0,sizeof(m_arr));
//i表示哪一行的元素，j表式哪一列的元素，i/3，j/3表示哪一块的元素
for(int i=0;i<9;i++){
for(int j=0;j<9;j++){
if(board[i][j]!='.'){
int temp = board[i][j]-'0'-1;
r_arr[i][temp]++; //把一行遍历完后遍历下一行
c_arr[j][temp]++; //每一列平均着遍历
m_arr[i/3][j/3][temp]++;
if(r_arr[i][temp]>1|| c_arr[j][temp]>1||m_arr[i/3][j/3][temp]>1){
return false;
}
}
}
}
return true;
}
};
