[转载]关于螺旋数组的面试题

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();

}