Dado un número N , la tarea es convertir cada dígito del número en palabras.
Ejemplos:
Entrada: N = 1234
Salida: Uno Dos Tres Cuatro
Explicación:
Cada dígito del número dado se ha convertido en su palabra correspondiente.
Entrada: N = 567
Salida: Cinco Seis Siete
Enfoque: La idea es recorrer cada dígito del número y usar switch-case . Dado que solo hay diez valores posibles para los dígitos, se pueden definir diez casos dentro de un bloque de cambio. Para cada dígito, se ejecutará su bloque de casos correspondiente y ese dígito se imprimirá en palabras.
A continuación se muestra la implementación del enfoque anterior:
CPP
// C++ implementation of the above approach
#include "bits/stdc++.h"
using namespace std;
// Function to return the word
// of the corresponding digit
void printValue(char digit)
{
// Switch block to check for each digit c
switch (digit) {
// For digit 0
case '0':
cout << "Zero ";
break;
// For digit 1
case '1':
cout << "One ";
break;
// For digit 2
case '2':
cout << "Two ";
break;
// For digit 3
case '3':
cout << "Three ";
break;
// For digit 4
case '4':
cout << "Four ";
break;
// For digit 5
case '5':
cout << "Five ";
break;
// For digit 6
case '6':
cout << "Six ";
break;
// For digit 7
case '7':
cout << "Seven ";
break;
// For digit 8
case '8':
cout << "Eight ";
break;
// For digit 9
case '9':
cout << "Nine ";
break;
}
}
// Function to iterate through every
// digit in the given number
void printWord(string N)
{
int i, length = N.length();
// Finding each digit of the number
for (i = 0; i < length; i++) {
// Print the digit in words
printValue(N[i]);
}
}
// Driver code
int main()
{
string N = "123";
printWord(N);
return 0;
}
Java
// Java implementation of the above approach
class GFG
{
// Function to return the word
// of the corresponding digit
static void printValue(char digit)
{
// Switch block to check for each digit c
switch (digit)
{
// For digit 0
case '0':
System.out.print("Zero ");
break;
// For digit 1
case '1':
System.out.print("One ");
break;
// For digit 2
case '2':
System.out.print("Two ");
break;
// For digit 3
case '3':
System.out.print("Three ");
break;
// For digit 4
case '4':
System.out.print("Four ");
break;
// For digit 5
case '5':
System.out.print("Five ");
break;
// For digit 6
case '6':
System.out.print("Six ");
break;
// For digit 7
case '7':
System.out.print("Seven ");
break;
// For digit 8
case '8':
System.out.print("Eight ");
break;
// For digit 9
case '9':
System.out.print("Nine ");
break;
}
}
// Function to iterate through every
// digit in the given number
static void printWord(String N)
{
int i, length = N.length();
// Finding each digit of the number
for (i = 0; i < length; i++)
{
// Print the digit in words
printValue(N.charAt(i));
}
}
// Driver code
public static void main(String[] args)
{
String N = "123";
printWord(N);
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 implementation of the above approach
# Function to return the word
# of the corresponding digit
def printValue(digit):
# Switch block to check for each digit c
# For digit 0
if digit == '0':
print("Zero ", end = " ")
# For digit 1
elif digit == '1':
print("One ", end = " ")
# For digit 2
elif digit == '2':
print("Two ", end = " ")
#For digit 3
elif digit=='3':
print("Three",end=" ")
# For digit 4
elif digit == '4':
print("Four ", end = " ")
# For digit 5
elif digit == '5':
print("Five ", end = " ")
# For digit 6
elif digit == '6':
print("Six ", end = " ")
# For digit 7
elif digit == '7':
print("Seven", end = " ")
# For digit 8
elif digit == '8':
print("Eight", end = " ")
# For digit 9
elif digit == '9':
print("Nine ", end = " ")
# Function to iterate through every
# digit in the given number
def printWord(N):
i = 0
length = len(N)
# Finding each digit of the number
while i < length:
# Print the digit in words
printValue(N[i])
i += 1
# Driver code
N = "123"
printWord(N)
# This code is contributed by mohit kumar 29
C#
// C# implementation of the above approach
using System;
class GFG
{
// Function to return the word
// of the corresponding digit
static void printValue(char digit)
{
// Switch block to check for each digit c
switch (digit)
{
// For digit 0
case '0':
Console.Write("Zero ");
break;
// For digit 1
case '1':
Console.Write("One ");
break;
// For digit 2
case '2':
Console.Write("Two ");
break;
// For digit 3
case '3':
Console.Write("Three ");
break;
// For digit 4
case '4':
Console.Write("Four ");
break;
// For digit 5
case '5':
Console.Write("Five ");
break;
// For digit 6
case '6':
Console.Write("Six ");
break;
// For digit 7
case '7':
Console.Write("Seven ");
break;
// For digit 8
case '8':
Console.Write("Eight ");
break;
// For digit 9
case '9':
Console.Write("Nine ");
break;
}
}
// Function to iterate through every
// digit in the given number
static void printWord(string N)
{
int i, length = N.Length;
// Finding each digit of the number
for (i = 0; i < length; i++)
{
// Print the digit in words
printValue(N[i]);
}
}
// Driver code
public static void Main()
{
string N = "123";
printWord(N);
}
}
// This code is contributed by AnkitRai01
Javascript
<script>
// JavaScript implementation of
// the above approach
// Function to return the word
// of the corresponding digit
function printValue(digit) {
// Switch block to check for
// each digit c
switch (digit) {
// For digit 0
case "0":
document.write("Zero ");
break;
// For digit 1
case "1":
document.write("One ");
break;
// For digit 2
case "2":
document.write("Two ");
break;
// For digit 3
case "3":
document.write("Three ");
break;
// For digit 4
case "4":
document.write("Four ");
break;
// For digit 5
case "5":
document.write("Five ");
break;
// For digit 6
case "6":
document.write("Six ");
break;
// For digit 7
case "7":
document.write("Seven ");
break;
// For digit 8
case "8":
document.write("Eight ");
break;
// For digit 9
case "9":
document.write("Nine ");
break;
}
}
// Function to iterate through every
// digit in the given number
function printWord(N) {
var i,
length = N.length;
// Finding each digit of the number
for (i = 0; i < length; i++) {
// Print the digit in words
printValue(N[i]);
}
}
// Driver code
var N = "123";
printWord(N);
</script>
Producción
One Two Three
Enfoque recursivo
Enfoque: la idea es llamar recursivamente a la función hasta que el número se convierta en cero dividiéndolo por 10 y almacenando el resto en una variable. Luego imprimimos el dígito de la array de strings, ya que el dígito almacenará el índice del número que se imprimirá de la array.
imprimimos el número después de la llamada recursiva para mantener el orden de los dígitos en el número de entrada, si imprimimos antes de la llamada de función recursiva, el nombre del dígito se imprimirá en orden inverso.
Como estamos dividiendo n entre 10 en cada llamada recursiva, la relación de recurrencia será T(n) = T(n/10) + 1
Complejidad de tiempo = O (log n)
A continuación se muestra la implementación del enfoque anterior:
C++
//C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
void ToDigits(int n, string arr[])
{
// base case
if (n == 0) {
return;
}
// storing the last digit of the number and updating
// number
int digit = n % 10;
n = n / 10;
// recursive call
ToDigits(n, arr);
// printing the digits form the string array storing name
// of the given index
cout << arr[digit] << " ";
}
int main()
{
string arr[10]
= { "zero", "one", "two", "three", "four",
"five", "six", "seven", "eight", "nine" };
int n;
n = 123; //it can be changed to take user input
ToDigits(n, arr);
return 0;
}
//This code is contributed by rahulpatel43433
Java
/*package whatever //do not write package name here */
import java.io.*;
class GFG {
static void ToDigits(int n, String[] arr)
{
// base case
if (n == 0) {
return;
}
// storing the last digit of the number and updating
// number
int digit = n % 10;
n = n / 10;
// recursive call
ToDigits(n, arr);
// printing the digits form the string array storing name
// of the given index
System.out.print(arr[digit]);
System.out.print(" ");
}
// Driver Code
public static void main(String args[])
{
String[] arr = { "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine" };
int n = 123; //it can be changed to take user input
ToDigits(n, arr);
}
}
// This code is contributed by shinjanpatra
Python3
# Python implementation of above approach def ToDigits(n, arr): # base case if (n == 0): return # storing the last digit of the number and updating # number digit = n % 10 n = n // 10 # recursive call ToDigits(n, arr) # printing the digits form the string array storing name # of the given index print(arr[digit] , end = " ") # driver code arr = [ "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine" ] n = 123 #it can be changed to take user input ToDigits(n, arr) # This code is contributed by shinjanpatra
C#
//c# implementation of above approach
using System;
public class GFG {
static void ToDigits(int n, String[] arr)
{
// base case
if (n == 0) {
return;
}
// storing the last digit of the number and updating
// number
int digit = n % 10;
n = n / 10;
// recursive call
ToDigits(n, arr);
// printing the digits form the string array storing name
// of the given index
Console.Write(arr[digit]+" ");
}
// Driver program to test above
public static void Main()
{
String[] arr = new string[10]{ "zero", "one", "two", "three", "four", "five",
"six", "seven", "eight", "nine" };
int n;
n = 123; //it can be changed to take user input
ToDigits(n, arr);
}
}
// This code is contributed by aditya942003patil
Javascript
<script>
// JavaScript implementation of above approach
function ToDigits(n, arr)
{
// base case
if (n == 0) {
return;
}
// storing the last digit of the number and updating
// number
let digit = n % 10;
n = Math.floor(n / 10);
// recursive call
ToDigits(n, arr);
// printing the digits form the string array storing name
// of the given index
document.write(arr[digit] , " ");
}
// driver code
let arr = [ "zero", "one", "two", "three", "four",
"five", "six", "seven", "eight", "nine" ]
let n = 123; //it can be changed to take user input
ToDigits(n, arr);
// This code is contributed by shinjanpatra
</script>
Producción
one two three
Artículo relacionado: Imprime dígitos individuales como palabras sin usar if o switch
Publicación traducida automáticamente
Artículo escrito por Akshita33Patwal y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA