2.1.14 —线性表—Valid Sudoku

最新推荐文章于 2020-12-27 21:35:36 发布

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本文介绍了一个用于验证数独是否有效的算法实现。该算法通过检查数独中的每一行、每一列以及每一个宫格内的数字是否重复来判断数独是否有效。代码采用C++编写，并提供了一个具体的数独示例进行验证。

描述
Determine if a Sudoku is valid, according to: Sudoku Puzzles - e Ruleshttp://sudoku.com.au/TheRules.aspx .

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

#include<iostream>
using namespace std;
const int COL = 9;
const int SUB = 3;
bool ValidSudoku(int a[][COL], int row)
{
	bool flag = false;
	for (int i = 0; i < row; i++)//处理行
	{
		bool temp[COL] = { false };
		for (int j = 0; j < COL; j++)
		{
			if (a[i][j] != 0)
			{
				if (!temp[a[i][j] - 1])
					temp[a[i][j] - 1] = true;
				else
					return flag;
			}
		}
	}
	//
	for (int j = 0; j < COL; j++)//处理列
	{
		bool temp[COL] = { false };
		for (int i = 0; i < row; i++)
		{
			if (a[i][j] != 0)
			{
				if (!temp[a[i][j] - 1])
					temp[a[i][j] - 1] = true;
				else
					return flag;
			}
		}
	}
	//处理每个子宫格
	for (int i = 0; i < row - 2; i++)
	{
		for (int j = 0; j < COL - 2; j++)
		{
			for (int k1 = i; k1 < i + 3; k1++)
			{
				bool temp[9] = { false };
				for (int k2 = j; k2 < j + 3; k2++)
				{
					if (a[k1][k2] != 0)
					{
						if (!temp[a[k1][k2] - 1])
							temp[a[k1][k2] - 1] = true;
						else
							return flag;
					}
				}
			}
			//
			for (int k1 = j; k1 < j + 3; k1++)
			{
				bool temp[9] = { false };
				for (int k2 = i; k2 < i + 3; k2++)
				{
					if (a[k2][k1] != 0)
					{
						if (!temp[a[k2][k1] - 1])
							temp[a[k2][k1] - 1] = true;
						else
							return flag;
					}
				}
			}
			//
		}
	}
	flag = true;
	return flag;
}
int main()
{
	int a[COL][COL] =
	{
		{ 5, 3, 0, 0, 7, 0, 0, 0, 0 }, { 6, 0, 0, 1, 9, 5, 0, 0, 0 }, { 0, 9, 8, 0, 0, 0, 0, 6, 0 },
		{ 8, 0, 0, 0, 6, 0, 0, 0, 3 }, { 4, 0, 0, 8, 0, 3, 0, 0, 1 }, { 7, 0, 0, 0, 2, 0, 0, 0, 6 },
		{ 0, 6, 0, 0, 0, 0, 2, 8, 0 }, { 0, 0, 0, 4, 1, 9, 0, 0, 5 }, { 0, 0, 0, 0, 8, 0, 0, 7, 9 },
	};
	bool flag = ValidSudoku(a, COL);
	if (flag)
		cout << "this is valid sudoku!" << endl;
}