原文:
给定一个方阵 mat[][] 和一个正整数 k 。任务是打印mat【】【】的 k 边界。
示例:
输入: 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}} 输出:1 2 3 4 5 6 10【t100
输入: mat[][] = {{1，2，3}，k = 2 {4，5，6} {7，8，9}} 输出: 5
方法:这个问题是基于实现的。遍历矩阵，检查每个元素是否位于 kth 边界上。如果是，则打印元素否则打印空格字符。按照以下步骤解决给定的问题。
- 对于我从 0 到 n
- 用于从 0 到 n 的 j
- if((i = = k–1 或 i = = n–k)和(j > = k–1 和 j < = n–k))
- 打印垫[i][j]
- 否则，如果(j = = k–1 或 j = = n–k)和(i > = k–1 和 i < = n–k):
- 打印垫[i][j]
- if((i = = k–1 或 i = = n–k)和(j > = k–1 和 j < = n–k))
- 用于从 0 到 n 的 j
- 这将给出 mat[][]所需的 kth 边框
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 语言(一种计算机语言，尤用于创建网站)
// 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.
python 3
# 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.
java 描述语言
output
1 2 3 4 5
6 10
11 15
16 20
21 22 23 24 25
时间复杂度:o(n^2) t3】空间复杂度: o(1)
麻将胡了pg电子网站的版权属于：月萌api www.moonapi.com，转载请注明出处
