《程序员面试宝典》之螺旋队列问题学习

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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tint u = t + t;              // u = 2tint v = u - 1;              // v = 2t-1v = v * v + u;              // v = (2t-1)^2 + 2tif(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;}

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