MxN螺旋矩阵(由外向内)

问题描述

按顺时针方向构建一个MxN的螺旋矩阵（或按顺时针方向螺旋访问一个MxN的矩阵）：在不构造螺旋矩阵的情况下，给定坐标i、j值求其对应的值f(i, j)。如对11x7 矩阵， f(4, 0) = 29 f(4, 1) = 54 f(4, 3) = 74 f(4, 4) = 61

解析

对MxN矩阵，最先访问最外层的MxN的矩形上的元素，接着再访问里面一层的(M-2)x(N-2)矩形上的元素.最后可能会剩下一些元素，组成一个点或一条线。对第i个矩阵（i=0,1,2.....）,4个顶点的坐标为：

(i,i) ------------------------------(i, n-1-i)

| |

(m-1-i)---------------------------(m-1-i,n-1-i)

要访问该矩形上的所有元素，只须用4个for循环，每个循环访问一个点和一边条边上的元素即可。另外，要注意对最终可能剩下的1xK或Kx1矩阵再做个特殊处理。代码：

inline void act(int t) { printf("%3d ", t);
}const int small = col<row?col:row;
const int count = small/2;for (int i = 0; i < count; ++i) {const int C = col-1-i;const int R = row-1-i;for (int j = i; j < C; ++j)act(arr[i][j]);for (int j = i; j < R; ++j) act(arr[j][C]);for (int j = C; j > i; --j) act(arr[R][j]);for (int j = R; j > i; --j) act(arr[j][i]);}if (small & 1) {const int i = count;if (row <= col){for (int j = i; j < col-i; ++j)act(arr[i][j]);else{for (int j = i; j < row-i; ++j) act(arr[j][i]);}
}

如果只是构建螺旋矩阵的话，稍微修改可以实现4个for循环独立：

const int small = col < row ? col : row;const int count = small / 2;for (int i = 0; i < count; ++i) {const int C = col-1-i;const int R = row-1-i;const int cc = C-i;const int rr = R-i;const int s = 2*i*(row+col-2*i)+1;for (int j = i, k = s; j < C; ++j) arr[i][j] = k++;for (int j = i, k = s+cc; j < R; ++j) arr[j][C] = k++;for (int j = C, k = s+cc+rr; j > i; --j) arr[R][j] = k++;for (int j = R, k = s + cc * 2 + rr; j > i; --j) arr[j][i] = k++;}if (small&1) {const int i = count;int k = 2*i*(row+col-2*i)+1;if (row <= col){for (int j = i; j <col-i; ++j) arr[i][j] = k++;} else{for (int j = i; j <row-i; ++j) arr[j][i] = k++;} }

关于s的初始值取2i(row+col-2*i)+1

由于C++的二维数组是通过一维数组实现的。二维数组的实现一般有下面三种：1)静态分配足够大的数组；2)动态分配一个长为m*n的一维数组；3)动态分配m个长为n的一维数组，并将它们的指针存在一个长为m的一维数组。二维数组的不同实现方法，对函数接口有很大影响。给定坐标直接求值f(x,y);如前面所述，对第i个矩形（i=0, 1, 2 …），4个顶点的坐标为：

(i,i) ------------------------------(i, n-1-i)

| |

(m-1-i)---------------------------(m-1-i,n-1-i)

对给定的坐标(x,y)，如果它落在某个这类矩形上，显然其所在的矩形编号为：k = min{ x,y,m-1-x,n-1-y};MxN矩阵删除访问第k个矩形前所访问的所有元素后，可得到(m-2*k)*(n-2*k)矩阵，因此已访问的元素个数为：m*n-(m-2*k)*(n-2*k)=2*k*(m+n-2*k)，因而(k,k)对应的值为：T(k) = 2*k*(m+n-2*k)+1.对某个矩形，设点(x, y)到起始点(k,k)的距离d = x-k + y-k = x+y-2*k;① 向右和向下都只是横坐标或纵坐标增加1，这两条边上的点满足f(x, y) = T(k) + d;② 向左和向下都只是横坐标或纵坐标减少1，这两条边上的点满足f(x, y) = T(k+1) - d.如果给定坐标的点(x, y)，不在任何矩形上，则它在一条线上，仍满足f(x, y) = T(k) + d

int getv(int row, int col, int max_row, int max_col){ // row < max_row, col < max_colint level, diatancce;int start_value, next_vlue;level = min(min(row, max_row - 1 - row), min(col, max_col - 1 - col));distance = row + col - level * 2;start_value = 2 * level * (max_row + max_col - 2 * level) + 1;if (row == level || col == max_col - 1 - level ||(max_col < max_row && level * 2 + 1 == max_col))return start_value + distance;next_value = start_value + (max_row + max_col - 4 * level - 2) * 2;return next_value - distance;
}

特别说明

上面的讨论都是基于MxN矩阵的，对于特例NxN矩阵，可以做更多的优化。比如构建螺旋矩阵，如果n为奇数，则矩阵可以拆分为几个矩形加上一个点。前面的条件判断可以优化为：if(small&i) act[count][count];甚至可以调整4个for循环的遍历元素个数（前面代码，每个for循环遍历n-1-2*i个元素，可以调整为：n-2*i，n-1-2*i, n-1-2*i,n-2-2*i）从而达到省略if判断。

代码：

//螺旋矩阵，给定坐标直接求值
#include<iostream>
#include<algorithm>using std::min;
using std::cout;/*
int getv2(int row, int col, int max_row, int max_col){ // row < max_row, col < max_colint level,distance;int start_value, next_value;level = min(min(row, max_row - 1 - row), min(col, max_col - 1 - col));distance = row + col - level * 2;start_value = 2 * level * (max_row + max_col - 2 * level) + 1;if (row == level || col == max_col - 1 - level) return(start_value + distance);//++level; int next_value = 2 * level * (max_row + max_col - 2 * level) + 1;next_value = start_value + (max_row + max_col - 4 * level - 2) * 2;if (next_value > max_col * max_row) return(start_value + distance);return (next_value - distance);
}*/int getv(int row, int col, int max_row, int max_col){ // row < max_row, col < max_colint level = min(min(row, max_row - 1 - row), min(col, max_col - 1 - col));int distance = row + col - level * 2;int start_value = 2 * level * (max_row + max_col - 2 * level) + 1;if (row == level || col == max_col - 1 - level || (max_col < max_row && level * 2 + 1 == max_col))return start_value + distance;//++level; int next_value = 2 * level * (max_row + max_col - 2 * level) + 1;int next_value = start_value + (max_row + max_col - 4 * level - 2) * 2;return next_value - distance;} int main( ){int test[][2] = {{5, 5}, {5, 7}, {7, 5}, {4, 4}, {4, 6}, {6, 4}};const int sz = sizeof(test) / sizeof(test[0]);for (int k = 0; k < sz; ++k) {int M = test[k][0];int N = test[k][1];  for (int i = 0; i < M; ++i) {for (int j = 0; j < N; ++j)cout.width(4), cout << getv(i, j, M, N) << " ";cout << "\n";}cout << "\n";}
}

完整的代码

//螺旋矩阵 #include<iostream>
int counter = 0;
inline void act(int& t){//std::cout.width(3), std::cout << t;t = ++::counter;
}void act_arr(int *arr, int row, int col, int max_col){ //col < max_colconst int small = col < row ? col : row;const int count = small / 2;int *p = arr;for (int i = 0; i < count; ++i) {const int C = col - 1 - 2 * i;const int R = row - 1 - 2 * i;for (int j = 0; j < C; ++j) act(*p++);for (int j = 0; j < R; ++j) act(*p), p += max_col;for (int j = 0; j < C; ++j) act(*p--);for (int j = 0; j < R; ++j) act(*p), p -= max_col;p += max_col + 1;}if (small & 1) {const int i = count;if (row <= col) for (int j = 0; j < col - 2 * i; ++j) act(*p++);else for (int j = 0; j < row - 2 * i; ++j) act(*p), p += max_col;}
} void act_arr(int* arr[], int row, int col){const int small = col < row ? col : row;const int count = small / 2;for (int i = 0; i < count; ++i) {const int C = col - 1 - i;const int R = row - 1 - i;for (int j = i; j < C; ++j) act(arr[i][j]);for (int j = i; j < R; ++j) act(arr[j][C]);for (int j = C; j > i; --j) act(arr[R][j]);for (int j = R; j > i; --j) act(arr[j][i]);}if (small & 1) {const int i = count;if (row <= col) for (int j = i; j < col - i; ++j) act(arr[i][j]);else for (int j = i; j < row - i; ++j) act(arr[j][i]);}
}void act_arr_2(int* arr[], int row, int col){const int small = col < row ? col : row;const int count = small / 2;for (int i = 0; i < count; ++i) {const int C = col - 1 - i;const int R = row - 1 - i;const int cc = C - i;const int rr = R - i;const int s = 2 * i * (row + col - 2 * i) + 1;for (int j = i, k = s; j < C; ++j) arr[i][j] = k++;for (int j = i, k = s + cc; j < R; ++j) arr[j][C] = k++;for (int j = C, k = s + cc + rr; j > i; --j) arr[R][j] = k++;for (int j = R, k = s + cc * 2 + rr; j > i; --j) arr[j][i] = k++;}
if (small & 1) {const int i = count;int k = 2 * i * (row + col - 2 * i) + 1;if (row <= col) for (int j = i; j < col - i; ++j) arr[i][j] = k++;else for (int j = i; j < row - i; ++j) arr[j][i] = k++;}
}void print_arr(int *arr, int row, int col, int max_col){ //col < max_colfor (int i = 0, *q = arr; i < row; ++i, q += max_col) {for (int *p = q; p < q + col; ++p)std::cout.width(4), std::cout << *p;std::cout << "\n";}std::cout << "\n";
}void print_arr(int* a[], int row, int col){ //col < max_colfor (int i = 0; i < row; ++i) {for (int j = 0; j < col; ++j)std::cout.width(4), std::cout << a[i][j];std::cout << "\n"; } std::cout << "\n";
}void test_1(){const int M = 25;const int N = 25;int a[M][N];int test[][2] = {{5, 5}, {5, 7}, {7, 5}, {4, 4}, {4, 6}, {6, 4}};const int sz = sizeof(test) / sizeof(test[0]);std::cout << "Test 1:\n";for (int i = 0; i < sz; ++i) {int row = test[i][0];int col = test[i][1];if (row < 0 || row > M) row = 3;if (col < 0 || col > N) col = 3;::counter = 0;act_arr(&a[0][0], row, col, N);print_arr(&a[0][0], row, col, N);}
}void test_2(){int test[][2] = {{5, 5}, {5, 7}, {7, 5}, {4, 4}, {4, 6}, {6, 4}};const int sz = sizeof(test) / sizeof(test[0]);std::cout << "Test 2:\n";for (int i = 0; i < sz; ++i) {int row = test[i][0];int col = test[i][1]; int **arr = new int*[row];for (int i = 0; i < row; ++i) arr[i] = new int[col];::counter = 0;act_arr(arr, row, col);print_arr(arr, row, col);for (int i = 0; i < row; ++i) delete[] arr[i];delete[] arr; }
}int main( ){test_1();test_2();
}

C语言经典NXN螺旋矩阵(由外向内)代码

//N阶螺旋矩阵
#include <stdio.h>
#include <stdlib.h>
int main()
{int N,i,j,n,num=1;int a[10][10]={0};printf("输入你要输出的几阶中断:");scanf("%d",&N);for(n=0;n<=N/2;n++){for(j=n;j<=N-n-1;j++)a[n][j]=num++;for(i=n+1;i<N-n-1;i++)a[i][N-n-1]=num++;for(j=N-n-1;j>n;j--)a[N-n-1][j]=num++;for(i=N-n-1;i>n;i--)a[i][n]=num++;}    //输出螺旋矩阵 for(i=0;i<N;i++){for(j=0;j<N;j++)printf("%2d ",a[i][j]);printf("\n");}    system("pause");return 0;
}

思路是这样的，刚开始很容易想到顺时针赋值，如下图为5阶：分为四个方向顺时针赋值。每个方向负责相同数量的元素。

但这样，后来发现当N为基数时，最中心一个数不能被赋值。所以改为还是顺时针赋值，只是->方向多负责一个元素，右| 方向少负责一个元素

输出结果为：

====================================================================

转载请注明出处：http://blog.csdn.net/utimes/article/details/8779143

====================================================================