打印矩陣、旋轉矩陣、矩陣查找（c++）

不能想如何進行座標位置變換到下一個，很難。
應該考慮數據狀況，從宏觀上進行分解。

1 轉圈打印矩陣

1.1 打印一空心圈

#include <iostream>
using namespace std;
void printH(int** a, int aR, int aC, int bR, int bC)
{
	int indexac = aC;
	int indexar = aR;
	int indexbc = bC;
	int indexbr = bR;
	while (aC<bC)
		cout << *((int*)a + (bC- indexac+1) * aR + aC++) << " ";	// 
	while (aR < bR)
		cout << *((int*)a + (bC - indexac + 1) * (aR++) + aC) << " ";	// 
	while (aC > indexac)
		cout << *((int*)a + (bC - indexac + 1) * aR + aC--) << " ";	// 
	while (aR > indexar)
		cout << *((int*)a + (bC - indexac + 1) * (aR--) + aC) << " ";	//
	cout << endl;
}

int main()
{
	int a[3][4] = { {1,2,3,4},{5,6,7,8},{9,10,11,12} };
	for (int i = 0; i < 3; i++)
		for (int j = 0; j < 3; j++)
			cout << a[i][j];
	cout << endl;
	printH((int**)a, 0, 0, 2, 3);
	system("pause");
	return 0;
}

1.2 打印函數

若傳入數組爲二維（int**）a，且不能確定長和寬。
way1：下面方法也能獲取長度

1 int lines = sizeof(a) / sizeof(a[0][0]);
2 int row = sizeof(a) / sizeof(a[0]);
3 int column = lines / row;

way2：f爲列數

int a[s][f];
 a[i][j] = *((int*)a + (f ) * i + j)

當二維數組不是在子函數中時，可以用1的方法求長度，在子函數中則不能求，可用vector方法。後續補上。

#include <iostream>
using namespace std;
void getlength(int** a)
{
	int lines = sizeof(a) / sizeof(*((int*)a));
	int row = sizeof(a) / sizeof((int*)a);
	int column = lines / row;
	cout << row << " " << column << endl;
}
int main()
{
	
	int a[3][3] = { {1,2,3},{3,4,5},{3,4,5} };

	//int lines = sizeof(a) / sizeof(a[0][0]);
	//int row = sizeof(a) / sizeof(a[0]);
	//int column = lines / row;
	//cout << row << " " << column << endl;

	getlength((int**)a);
	system("pause");
	return 0;
}

#include <iostream>
using namespace std;
//f爲求二維動態數組的值
void printH(int** a, int aR, int aC, int bR, int bC,int f)
{
	int indexac = aC;
	int indexar = aR;
	int indexbc = bC;
	int indexbr = bR;

	if (indexar == indexbr)
		while (aC <= bC)
			cout << *((int*)a + (f + 1) * aR + aC++) << " ";	// 
	else if (indexac == indexbc)
		while (aR <= bR)
			cout << *((int*)a + (f + 1) * (aR++) + aC) << " ";
	else
	{
		while (aC < bC)
			cout << *((int*)a + (f + 1) * aR + aC++) << " ";	// 
		while (aR < bR)
			cout << *((int*)a + (f + 1) * (aR++) + aC) << " ";	// 
		while (aC > indexac)
			cout << *((int*)a + (f + 1) * aR + aC--) << " ";	// 
		while (aR > indexar)
			cout << *((int*)a + (f + 1) * (aR--) + aC) << " ";	//
		//cout << endl;
	}
}
//a[5][5],則輸入i爲4，j爲4
void printAll(int** a, int i, int j)
{
	int ac = 0, ar = 0;
	int bc = j, br = i;
	while (ac <= bc && ar <= br)
		printH(a, ar++, ac++, br--, bc--,j);
}
int main()
{
	int a[5][5] = { {1,2,3,4,5},{6,7,8,9,10},{11,12,13,14,15},{16,17,18,19,20},{21,22,23,24,25} };
	for (int i = 0; i < 5; i++)
		for (int j = 0; j < 5; j++)
			cout << a[i][j];
	cout << endl;
	//printH((int**)a, 0, 0, 2, 3);
	printAll((int**)a, 4, 4);
	system("pause");
	return 0;
}

2 旋轉正方形矩陣：vector方法（後續更新）

3 之打印（後續更新）

4 矩陣查找（後續更新）