打印矩阵的第 K 个边界

给定一个方阵mat[][]和一个正整数K 。任务是打印mat[][]的第K个边界。

例子：

Input: mat[][] = {{1, 2, 3, 4, 5}, K = 1
{6, 7, 8, 9, 10}
{11, 12, 13, 14, 15}
{16, 17, 18, 19, 20}
{21, 22, 23, 24, 25}}
Output: 1 2 3 4 5
6 10
11 15
16 20
21 22 23 24 25
Explanation: The first boundary of mat[][] is above.

Input: mat[][] = {{1, 2, 3}, K = 2
{4, 5, 6}
{7, 8, 9}}
Output: 5

编程需要懂一点英语

方法：这个问题是基于实现的。遍历矩阵并检查每个元素是否位于第 K 个边界上。如果是，则打印元素，否则打印空格字符。请按照以下步骤解决给定的问题。

对于i从0到N
- 对于j in 从0到N
  - if((i == K – 1 or i == N – K) and (j >= K – 1 and j <= N – K))
    - 打印垫[i][j]
  - else if (j == K – 1 or j == N – K) and (i >= K – 1 and i <= N – K):
    - 打印垫[i][j]
这将给出 mat[][] 所需的第 K 个边界

下面是上述方法的实现。

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to print Kth border of a matrix
void printKthBorder(vector> mat, int N, int K)
{
    for (int i = 0; i < N; i++)
    {
        cout << endl;
        for (int j = 0; j < N; j++)
        {
            // To keep track of which element to skip
            int flag = 0;
 
            if ((i == K - 1 || i == N - K) &&
                (j >= K - 1 && j <= N - K)) {
 
                // Print the element
                cout << mat[i][j] << " ";
 
                flag = 1;
            }
            else if ((j == K - 1 || j == N - K) &&
                    (i >= K - 1 && i <= N - K)) {
 
                // Print the element
                cout << mat[i][j] << " ";
 
                flag = 1;
            }
            if (flag == 0)
                cout << "  ";
        }
    }
}
 
// Driver code
int main() {
    int N = 5;
    int K = 1;
 
    vector> mat = {{1, 2, 3, 4, 5},
                            {6, 7, 8, 9, 10},
                            {11, 12, 13, 14, 15},
                            {16, 17, 18, 19, 20},
                            {21, 22, 23, 24, 25}};
 
    printKthBorder(mat, N, K);
}
 
// This code is contributed by Samim Hossain Mondal.

Java

// Java Program to implement
// the above approach
import java.util.*;
 
public class GFG
{
// Function to print Kth border of a matrix
static void printKthBorder(int [][]mat, int N, int K)
{
    for (int i = 0; i < N; i++)
    {
        System.out.println();
        for (int j = 0; j < N; j++)
        {
            // To keep track of which element to skip
            int flag = 0;
 
            if ((i == K - 1 || i == N - K) &&
                (j >= K - 1 && j <= N - K)) {
 
                // Print the element
                System.out.print(mat[i][j] + " ");
 
                flag = 1;
            }
            else if ((j == K - 1 || j == N - K) &&
                    (i >= K - 1 && i <= N - K)) {
 
                // Print the element
                System.out.print(mat[i][j] + " ");
 
                flag = 1;
            }
            if (flag == 0)
                System.out.print("  ");
        }
    }
}
 
// Driver code
public static void main(String args[]) {
    int N = 5;
    int K = 1;
 
    int [][]mat = {{1, 2, 3, 4, 5},
                    {6, 7, 8, 9, 10},
                    {11, 12, 13, 14, 15},
                    {16, 17, 18, 19, 20},
                    {21, 22, 23, 24, 25}};
 
    printKthBorder(mat, N, K);
}
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python program for above approach
 
# Function to print Kth border of a matrix
def printKthBorder(mat, N, K):
    for i in range(N):
        print()
        for j in range(N):
           
            # To keep track of which element to skip
            flag = 0
 
            if((i == K-1 or i == N-K) \
                  and (j >= K-1 and j <= N-K)):
               
                # Print the element
                print(mat[i][j], end =" ")
                 
                flag = 1
 
            elif (j == K-1 or j == N-K) \
                  and (i >= K-1 and i <= N-K):
               
                # Print the element
                print(mat[i][j], end =" ")
                 
                flag = 1
                 
            if flag == 0:
                print(end ="  ")
 
# Driver code
N = 5
K = 1
 
mat = [[1, 2, 3, 4, 5], \
       [6, 7, 8, 9, 10], \
       [11, 12, 13, 14, 15],
       [16, 17, 18, 19, 20], \
       [21, 22, 23, 24, 25]]
 
printKthBorder(mat, N, K)

C#

// C# Program to implement
// the above approach
using System;
 
class GFG
{
   
// Function to print Kth border of a matrix
static void printKthBorder(int [,]mat, int N, int K)
{
    for (int i = 0; i < N; i++)
    {
        Console.WriteLine();
        for (int j = 0; j < N; j++)
        {
            // To keep track of which element to skip
            int flag = 0;
 
            if ((i == K - 1 || i == N - K) &&
                (j >= K - 1 && j <= N - K)) {
 
                // Print the element
                Console.Write(mat[i, j] + " ");
 
                flag = 1;
            }
            else if ((j == K - 1 || j == N - K) &&
                    (i >= K - 1 && i <= N - K)) {
 
                // Print the element
                Console.Write(mat[i, j] + " ");
 
                flag = 1;
            }
            if (flag == 0)
                Console.Write("  ");
        }
    }
}
 
// Driver code
public static void Main() {
    int N = 5;
    int K = 1;
 
    int [,]mat = {{1, 2, 3, 4, 5},
                    {6, 7, 8, 9, 10},
                    {11, 12, 13, 14, 15},
                    {16, 17, 18, 19, 20},
                    {21, 22, 23, 24, 25}};
 
    printKthBorder(mat, N, K);
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

输出

1   2  3  4  5 
6           10 
11          15 
16          20 
21 22 23 24 25

时间复杂度： O(N^2)
空间复杂度： O(1)