问题
1 按顺时针方向构建一个m * n的螺旋矩阵（或按顺时针方向螺旋访问一个m * n的矩阵）：
2 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)。
比如对11 * 7矩阵， f(6, 0) = 27 f(6, 1) = 52 f(6, 3) = 76 f(6, 4) = 63
构建螺旋矩阵
对m * n 矩阵，最先访问最外层的m * n的矩形上的元素，接着再访问里面一层的 (m - 2) * (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（见图1）。
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, 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（见图1）。另外，要注意对最终可能剩下的1 * k 或 k * 1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 2 * i * (row + col - 2 * i) + 1请参考下一节。
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：
静态分配足够大的数组；
动态分配一个长为m*n的一维数组；
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, 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}
m*n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_col
{
int 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;
int 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m*n矩阵的，对于特例n*n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 & 1) 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。
测试代码
代码1：
//螺旋矩阵，给定坐标直接求值 by flyinghearts
//www.cnblogs.com/flyinghearts
#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_col
{
int 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) 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;
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_col
{
int 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";
}
}
代码2：
//螺旋矩阵 by flyinghearts#qq.com
//www.cnblogs.com/flyinghearts
#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_col
{
const 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_col
{
for (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_col
{
for (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();
}