Dado un número entero N , la tarea es imprimir el patrón del alfabeto H como se indica a continuación:
1 N 2 * 3 3 * 2 N * 3 2 1 * 2 3 3 2 * 1 N
Ejemplos:
Input: N = 3 Output: 1 3 2 2 3 2 1 2 2 1 3 Input: N = 4 Output: 1 4 2 3 3 2 4 3 2 1 3 2 2 3 1 4
Acercarse:
- Imprima el valor de la izquierda y deje 2 * (índice – 1) espacios en blanco e imprima el valor de la derecha.
- Imprime la N-ésima fila con N a 1 número.
- Repita el paso uno para (2 * N) – 1 vez para imprimir el patrón H deseado.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to print the desired
// Alphabet H Pattern
void alphabetPattern(int N)
{
// Declaring the values of left,
// middle, right side
int left = 0, middle = N - 1, right = N + 1;
// Main Row Loop
for (int row = 0; row < 2 * N - 1; row++) {
// Condition for the left Values
if (row < N)
cout << ++left;
else
cout << --left;
// Loop for the middle values
for (int col = 1; col < N - 1; col++) {
// Condition for the middleValues
if (row != N - 1)
// Two spaces for perfect alignment
cout << " "
<< " ";
else
cout << " " << middle--;
}
// Condition for the right Values
if (row < N)
cout << " " << --right;
else
cout << " " << ++right;
cout << endl;
}
}
// Driver Code
int main()
{
// Size of the Pattern
int N = 4;
alphabetPattern(N);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to print the desired
// Alphabet H Pattern
static void alphabetPattern(int N)
{
// Declaring the values of left,
// middle, right side
int left = 0, middle = N - 1, right = N + 1;
// Main Row Loop
for (int row = 0; row < 2 * N - 1; row++) {
// Condition for the left Values
if (row < N)
System.out.print( ++left);
else
System.out.print(--left);
// Loop for the middle values
for (int col = 1; col < N - 1; col++) {
// Condition for the middleValues
if (row != N - 1)
// Two spaces for perfect alignment
System.out.print( " ");
else
System.out.print( " " +middle--);
}
// Condition for the right Values
if (row < N)
System.out.print( " " +--right);
else
System.out.print( " " + ++right);
System.out.println();
}
}
// Driver Code
public static void main(String[] args) {
// Size of the Pattern
int N = 4;
alphabetPattern(N);
// This code is contributed by Rajput-Ji
}
}
Python3
# Python3 implementation of the approach
# Function to print the desired
# Alphabet H Pattern
def alphabetPattern(N):
# Declaring the values of left,
# middle, right side
left, middle, right = 0, N - 1, N + 1
# Main Row Loop
for row in range(0, 2 * N - 1):
# Condition for the left Values
if row < N:
left += 1
print(left, end = "")
else:
left -= 1
print(left, end = "")
# Loop for the middle values
for col in range(1, N - 1):
# Condition for the middleValues
if row != N - 1:
# Two spaces for perfect alignment
print(" ", end = " ")
else:
print(" " + str(middle), end = "")
middle -= 1
# Condition for the right Values
if row < N:
right -= 1
print(" " + str(right), end = "")
else:
right += 1
print(" " + str(right), end = "")
print()
# Driver Code
if __name__ == "__main__":
# Size of the Pattern
N = 4
alphabetPattern(N)
# This code is contributed by Rituraj Jain
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to print the desired
// Alphabet H Pattern
static void alphabetPattern(int N)
{
// Declaring the values of left,
// middle, right side
int left = 0, middle = N - 1, right = N + 1;
// Main Row Loop
for (int row = 0; row < 2 * N - 1; row++)
{
// Condition for the left Values
if (row < N)
Console.Write( ++left);
else
Console.Write(--left);
// Loop for the middle values
for (int col = 1; col < N - 1; col++)
{
// Condition for the middleValues
if (row != N - 1)
// Two spaces for perfect alignment
Console.Write( " ");
else
Console.Write( " " + middle--);
}
// Condition for the right Values
if (row < N)
Console.Write( " " + --right);
else
Console.Write( " " + ++right);
Console.WriteLine();
}
}
// Driver Code
public static void Main(String[] args)
{
// Size of the Pattern
int N = 4;
alphabetPattern(N);
}
}
// This code is contributed by
// PrinciRaj1992
PHP
<?php
// PHP implementation of the approach
// Function to print the desired
// Alphabet H Pattern
function alphabetPattern($N)
{
// Declaring the values of left,
// middle, right side
$left = 0;
$middle = $N - 1;
$right = $N + 1;
// Main Row Loop
for ($row = 0; $row < 2 * $N - 1; $row++)
{
// Condition for the left Values
if ($row < $N)
echo (++$left);
else
echo (--$left);
// Loop for the middle values
for ($col = 1; $col < $N - 1; $col++)
{
// Condition for the middleValues
if ($row != $N - 1)
// Two spaces for perfect alignment
echo " "." ";
else
echo " ".($middle--);
}
// Condition for the right Values
if ($row < $N)
echo " ".(--$right);
else
echo " ".(++$right);
echo "\n";
}
}
// Driver Code
// Size of the Pattern
$N = 4;
alphabetPattern($N);
// This code is contributed by mits
?>
Javascript
<script>
// JavaScript implementation
// of the approach
// Function to print the desired
// Alphabet H Pattern
function alphabetPattern(N)
{
// Declaring the values of left,
// middle, right side
var left = 0,
middle = N - 1,
right = N + 1;
// Main Row Loop
for (var row = 0; row < 2 * N - 1; row++)
{
// Condition for the left Values
if (row < N) {
++left;
document.write(left);
} else {
--left;
document.write(left);
}
// Loop for the middle values
for (var col = 1; col < N - 1; col++)
{
// Condition for the middleValues
if (row != N - 1)
// Two spaces for perfect alignment
document.write(" " + " ");
else {
document.write(" " + middle);
middle--;
}
}
// Condition for the right Values
if (row < N) {
--right;
document.write(" " + right);
} else {
++right;
document.write(" " + right);
}
document.write("<br>");
}
}
// Driver Code
// Size of the Pattern
var N = 4;
alphabetPattern(N);
</script>
Producción:
1 4 2 3 3 2 4 3 2 1 3 2 2 3 1 4
Publicación traducida automáticamente
Artículo escrito por VISHAL_PERIYASAMY_R y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA