LeetCode算法 —— 有效的数独

题目：
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。

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。

---

代码如下所示：

#include 
#include 
#include 

using namespace std;
  
class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        for (size_t i = 0; i < 9; i++)
        {
            for (size_t j = 0; j < 9; j++)
            {
                if (board[i][j] != '.') {
                    bool isValid = JudgeIsNoWant(board, i * 9 + j);

                    if (!isValid) {
                        return false;
                    }
                }
            }
        }

        return true;
    }

private:
    bool JudgeIsNoWant(vector<vector<char>>& arr, int n)	// n 表示是当前的第几个数
    {
        int x = n / 9;  		// 定位当前数的位置（数据在二维数组中的位置）
        int y = n % 9;

        for (size_t i = 0; i < 9; i++)  // 横竖 搜索
        {
            if (arr[x][i] == arr[x][y] && i != y) return false;
            if (arr[i][y] == arr[x][y] && i != x) return false;
        }

        // 九宫格 搜索	这里需要一个小小的算法，确定九宫格的位置
        for (size_t i = x / 3 * 3; i < x / 3 * 3 + 3; i++)
            for (size_t j = y / 3 * 3; j < y / 3 * 3 + 3; j++)
                if (arr[i][j] == arr[x][y] && (i != x || j != y))
                    return false;

        return true;
    }
};



 

int main() {
	 
    /*
  世界上最难的数独:
    0 0 5 3 0 0 0 0 0
    8 0 0 0 0 0 0 2 0
    0 7 0 0 1 0 5 0 0
    4 0 0 0 0 5 3 0 0
    0 1 0 0 7 0 0 0 6
    0 0 3 2 0 0 0 8 0
    0 6 0 5 0 0 0 0 9
    0 0 4 0 0 0 0 3 0
    0 0 0 0 0 9 7 0 0
 */
   

	return 0;
}