LeetCode刷题--- 有效的数独

2023-12-28 11:54:18

个人主页:元清加油_【C++】,【C语言】,【数据结构与算法】-CSDN博客

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力扣递归算法题

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前言:这个专栏主要讲述递归递归、搜索与回溯剪枝算法,所以下面题目主要也是这些算法做的 ?

我讲述题目会把讲解部分分为3个部分:
1、题目解析

2、算法原理思路讲解

3、代码实现


有效的数独

题目链接:有效的数独

题目

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。

提示:

  • board.length == 9
  • board[i].length == 9
  • board[i][j]?是一位数字(1-9)或者?'.'

解法

算法原理思路讲解?

题目解法非常的简单,使用一个剪枝以及一个哈希策略即可

bool checkRow[9][10];
bool checkCol[9][10];
bool gird[3][3][10];
  • checkRow(用于判断行是否有重复)
  • checkCol(用于判断列是否有重复)
  • gird(用于判断 3x3 是否有重复)

以上思路讲解完毕,大家可以自己做一下了


代码实现

class Solution {
public:
bool checkRow[9][10];
bool checkCol[9][10];
bool gird[3][3][10];

bool isValidSudoku(vector<vector<char>>& board) 
{

	for (int i = 0; i < board.size(); i++)
	{
		for (int j = 0; j < board[i].size(); j++)
		{
			if (board[i][j] != '.')
			{
				int number = board[i][j] - '0';
				if (checkRow[i][number] || checkCol[j][number] || gird[i / 3][j / 3][number])
				{
					return false;
				}
				checkRow[i][number] = checkCol[j][number] = gird[i / 3][j / 3][number] = true;
			}
		}
	}
	return true;
}
};

文章来源:https://blog.csdn.net/weixin_74268082/article/details/135260743
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