36. 有效的数独

请你判断一个 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。（请参考示例图）
数独部分空格内已填入了数字，空白格用 '.' 表示。

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。
 

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。

 

 

#include<iostream>
#include<stack>
#include<algorithm>
#include<string>
#include<vector>
using namespace std;
/*
由于board中的整数限定在1到9的范围内，因此可以分别建立哈希表来存储任一个数在相应维度上是否出现过。
维度有3个：所在的行，所在的列，所在的box，注意box的下标也是从左往右、从上往下的。
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？
*/
class Solution {
public:
	bool isValidSudoku(vector<vector<char>>& board) {
		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
		// 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。
		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
		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。
		for (int i = 0; i<9; i++)
		{
			for (int j = 0; j<9; j++)
			{
				// 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，
				// 同时判断这个数在其所在的列有没有出现过
				// 同时判断这个数在其所在的box中有没有出现过
				if (board[i][j] == '.')
				{
					continue;
				}	
				int curNumber = board[i][j] - '0';
				if (row[i][curNumber])
				{
					return false;
				}
				if (col[j][curNumber])
				{
					return false;
				}
				if (box[j / 3 + (i / 3) * 3][curNumber])
				{
					return false;
				}
				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。
				col[j][curNumber] = 1;
				box[j / 3 + (i / 3) * 3][curNumber] = 1;
			}
		}
		return true;
	}
};