LeetCode:有效的数独

判断一个 9x9 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。


上图是一个部分填充的有效的数独。

数独部分空格内已填入了数字,空白格用 '.' 表示。

示例 1:

输入:
[
["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:
[
  ["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。
说明:

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。
给定数独序列只包含数字 1-9 和字符 '.' 。
给定数独永远是 9x9 形式的。

关键是怎么判断9个小数独,这里用公式(i/3)*3+j/3就可以计算出当前数所在的小数独。

 1 class Solution {
 2 public:
 3     bool isValidSudoku(vector<vector<char>>& board) {
 4         int row[9][10] = {0};// 哈希表存储每一行的每个数是否出现过,默认初始情况下,每一行每一个数都没有出现过
 5         // 整个board有9行,第二维的维数10是为了让下标有9,和数独中的数字9对应。
 6         int col[9][10] = {0};// 存储每一列的每个数是否出现过,默认初始情况下,每一列的每一个数都没有出现过
 7         int box[9][10] = {0};// 存储每一个box的每个数是否出现过,默认初始情况下,在每个box中,每个数都没有出现过。整个board有9个box。
 8         for(int i=0; i<9; i++){
 9             for(int j = 0; j<9; j++){
10                 // 遍历到第i行第j列的那个数,我们要判断这个数在其所在的行有没有出现过,
11                 // 同时判断这个数在其所在的列有没有出现过
12                 // 同时判断这个数在其所在的box中有没有出现过
13                 if(board[i][j] == '.') continue;
14                 int curNumber = board[i][j]-'0';
15                 if(row[i][curNumber]) return false; 
16                 if(col[j][curNumber]) return false;
17                 if(box[j/3 + (i/3)*3][curNumber]) return false;
18 
19                 row[i][curNumber] = 1;// 之前都没出现过,现在出现了,就给它置为1,下次再遇见就能够直接返回false了。
20                 col[j][curNumber] = 1;
21                 box[j/3 + (i/3)*3][curNumber] = 1;
22             }
23         }
24         return true;
25     }
26 };
27 
28 作者:liujin-4
29 链接:https://leetcode-cn.com/problems/valid-sudoku/solution/36-jiu-an-zhao-cong-zuo-wang-you-cong-shang-wang-x/
30 来源:力扣(LeetCode)
31 著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。

用数组直接记录,占用空间也比map或者set要小。

原文地址:https://www.cnblogs.com/rookiez/p/13203203.html