螺旋矩阵

螺旋矩阵

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）。

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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) {

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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：

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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++;

}

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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)

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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

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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 - co

l));

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;

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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

///flyinghearts

#include

#include

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using std::min;

using std::cout;

/*

int getv2(int row, int col, int max_row, int max_col) // row < max_row, co

l < 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 + distanc

e;

//++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;

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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 - co

l));

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;

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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)

(4), cout << getv(i, j, M, N) << " ";

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cout << "n";

}

cout << "n";

}

}

代码2：

//螺旋矩阵 by flyinghearts#

///flyinghearts

#include

int counter = 0;

inline void act(int& t)

{

//std::(3), std::cout << t;

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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--);

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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;

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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]);

}

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}

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++;

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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) {

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for (int *p = q; p < q + col; ++p)

std::(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::(4), std::cout << a[i][j];

std::cout << "n";

}

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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];

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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) {

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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()

{

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test_1();

test_2();

}

作者： flyinghearts

出处： /flyinghearts/

本文采用知识共享署名-非商业性使用-相同方式共享 2.5 中国大陆许可协议进行许

可，欢迎转载，但未经作者同意必须保留此段声明，且在文章页面明显位置给出原文连接，

否则保留追究法律责任的权利。

方阵填数

答案：

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#include

using namespace std;

int a[10][10];

void Fun(int n)

{

int m = 1,j,i;

for(i = 0; i < n/2; i++)

{

for(j = 0; j < n-i; j++)

{

if(a[i][j] == 0)

a[i][j] = m++;

}

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for(j = i+1; j < n-i; j++)

{

if(a[j][n-1-i] == 0)

a[j][n-1-i] = m++;

}

for(j = n-i-1; j > i; j--)

{

if(a[n-i-1][j] == 0)

a[n-i-1][j] = m++;

}

for(j = n-i-1; j > i; j--)

{

if(a[j][i] == 0)

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a[j][i] = m++;

}

}

if(n%2==1)

a[n/2][n/2] = m;

}

void main()

{

int n, i, j;

cin>>n;

for(int i = 0; i < n; i++)

{

for(int j = 0; j < n; j++)

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a[i][j] = 0;

}

Fun(n);

for(i = 0; i < n; i++)

{

for(int j = 0; j < n; j++)

{

cout << a[i][j] << " ";

}

cout <

}

}

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