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 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 <bits/stdc++.h>
using namespace std;
// Function to print Kth border of a matrix
void printKthBorder(vector<vector<int>> 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<vector<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 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
<script>
// JavaScript Program to implement
// the above approach
// Function to print Kth border of a matrix
function printKthBorder(mat, N, K)
{
for (let i = 0; i < N; i++)
{
document.write('<br>')
for (let j = 0; j < N; j++)
{
// To keep track of which element to skip
flag = 0
if ((i == K - 1 || i == N - K) &&
(j >= K - 1 && j <= N - K)) {
// Print the element
document.write(mat[i][j] + " ")
flag = 1
}
else if ((j == K - 1 || j == N - K) &&
(i >= K - 1 && i <= N - K)) {
// Print the element
document.write(mat[i][j] + " ")
flag = 1
}
if (flag == 0)
document.write(" ");
}
}
}
// 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)
// This code is contributed by Potta Lokesh
</script>
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)
版权属于：月萌API www.moonapi.com，转载请注明出处
