《程序员面试宝典》之螺旋队列问题学习

1.由内向外螺旋队列查找指定位置数值问题。

#include <iostream>
#include <stdio.h>

using namespace std;

#define max(a,b) (a < b ? b : a)
#define abs(a)   (a > 0 ? a : -a)

int MatrixCircle(int x, int y)
{
	/************************************************************************
	*           |北|上            
	*            |                                7  8  9
        * 西|左 ————————————>x(+) 东|右   6  1  2       
	*            |0(0,0)                          5  4  3
	*            |
	*           \|/y(+) 南|下        注明: 默认起始坐标(0,0)对应数值1
	*                                      x表示列, y表示行
	************************************************************************/


	int t = max(abs(x), abs(y));// 确定所在层t
	int u = t + t;              // u = 2t
	int v = u - 1;              // v = 2t-1
	v = v * v + u;              // v = (2t-1)^2 + 2t

	if(x == -t)                 // 队列增长方向与Y轴反向, 正西方向(y=0)数值(2t-1)^2+5t, v = (2t-1)^2+5t-y
	{
		v += u + t -y;          // v = (2t-1)^2 + 2t + 2t + t - y = (2t-1)^2 + 5t - y
	}
	else 
		if(y == -t)             // 队列增长方向与X轴同向, 正北方向(x=0)数值(2t-1)^2+7t, v = (2t-1)^2+7t+x
	    {
		  v += 3*u + x - t;     // v = (2t-1)^2 + 2t + 3*2t + x - t = (2t-1)^2 + 7t + x
	    }
		else 
		if(y == t)              // 队列增长方向与X轴反向, 正南方向(x=0)数值(2t-1)^2+3t, v = (2t-1)^2+3t-x
	    {
		   v += t - x;          // v = (2t-1)^2 + 2t + t - x = (2t-1)^2 + 3t - x
	    }
	    else                    // 队列增长方向与Y轴同向, 正东方向(y=0)数值(2t-1)^2+t, v = (2t-1)^2+t+y
	    {
		   v += y - t;          // v = (2t-1)^2 + 2t + y - t = (2t-1)^2 + t + y
	    }

	return v;
}

int main()
{
	cout << "Expection 7: " << MatrixCircle(-1, -1) << endl;
	cout << "Expection 2: " << MatrixCircle(1, 0) << endl;
	cout << "Expection 3: " << MatrixCircle(1, 1) << endl;
	cout << "Expection 1: " << MatrixCircle(0, 0) << endl;

	return 0;
}

2.顺时针递增矩阵打印

#include <iostream>

using namespace  std;

int a[10][10] = {0};

void OuterCircle(int n)
{
	int m = 1, j, i;
	
	for (i = 0; i < n/2; i++)
	{
		for (j = 0; j < n-i; j++) // 左-->右:每次顺时针循环的左上角到右上角
		{
			if(a[i][j] == 0)
			a[i][j] = m++;
		}

		for (j = i + 1; j < n-i-1; j++)// 上-->下:每次顺时针循环的右上角到右下角
		{
			if(a[j][n-1-i] == 0)
			a[j][n-1-i] = m++;
		}

		for (j = n-i-1; j > i; j--)// 右-->左:每次顺时针循环的右下角到左下角
		{
			if(a[n-i-1][j] == 0)
			a[n-i-1][j] = m++;
		}

		for (j = n-i-1; j > i; j--)// 下-->上:每次顺时针循环的左下角到右上角
		{
			if(a[j][i] == 0)
			a[j][i] = m++;
		}
	}
	// n为奇数时,最后顺时针递增矩阵中心只剩一个元素
	if (n%2 == 1)
	{
		a[n/2][n/2] = m;
	}
}
int main()
{
	cout << "Please input the int number for NxN matrix outercircle: " << endl;
	int n;
	cin >> n;
	cout << endl;

	// NxN矩阵由外向内顺时针循环打印矩阵值!
	OuterCircle(n);

	for (int j = 0; j < n; j++)
	{
		for(int i = 0; i < n; i++)
		{
			cout << a[j][i] << "\t";
		}
		cout << endl;
	}

	return 0;
}