Dados dos arreglos de enteros positivos A y B de tamaños M y N respectivamente, la tarea es empujar A[i] + B[i] en un nuevo arreglo para cada i = 0 a min(M, N) e imprimir el nuevo array generada al final. Si la suma es un número de dos dígitos, divida los dígitos en dos elementos, es decir, cada elemento de la array resultante debe ser un número de un solo dígito.
Ejemplos:
Entrada: A = {2, 3, 4, 5}, B = {1, 12, 3}
Salida: 3 1 5 7 5
2 + 1 = 3
3 + 12 = 15 = 1 5
4 + 3 = 7
5
Por lo tanto la array resultante será {3, 1, 5, 7, 5}
Entrada: A = {23, 5, 2, 7, 87}, B = {4, 67, 2, 8}
Salida: 2 7 7 2 4 1 5 8 7
Enfoque: Cree un vector para almacenar el resultado de cada suma. Si la suma es un número de un solo dígito, empuje el número en el vector; de lo contrario, divida el número en diferentes dígitos y empuje los dígitos en la array uno por uno. Imprime el contenido del vector al final.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <iostream>
#include <vector>
using namespace std;
// Utility function to print the contents of the vector
void printVector(vector<int>& result)
{
for (int i : result)
cout << i << " ";
}
// Recursive function to separate the digits of a positive
// integer and add them to the given vector
void split_number(int num, vector<int>& result)
{
if (num > 0) {
split_number(num / 10, result);
result.push_back(num % 10);
}
}
// Function to add two arrays
void add(vector<int> a, vector<int> b)
{
// Vector to store the output
vector<int> result;
int m = a.size(), n = b.size();
// Loop till a or b runs out
int i = 0;
while (i < m && i < n) {
// Get sum of next element from each array
int sum = a[i] + b[i];
// Separate the digits of sum and add them to
// the resultant vector
split_number(sum, result);
i++;
}
// Process remaining elements of first vector, if any
while (i < m) {
split_number(a[i++], result);
}
// Process remaining elements of second vector, if any
while (i < n) {
split_number(b[i++], result);
}
// Print the resultant vector
printVector(result);
}
// Driver code
int main()
{
// input vectors
vector<int> a = { 23, 5, 2, 7, 87 };
vector<int> b = { 4, 67, 2, 8 };
add(a, b);
return 0;
}
Java
// Java implementation of the above approach
import java.util.*;
class GFG
{
// Utility function to print
// the contents of the vector
static void printVector(Vector<Integer> result)
{
for (int i : result)
{
System.out.print(i + " ");
}
}
// Recursive function to separate
// the digits of a positive integer
// and add them to the given vector
static void split_number(int num, Vector<Integer> result)
{
if (num > 0)
{
split_number(num / 10, result);
result.add(num % 10);
}
}
// Function to add two arrays
static void add(Vector<Integer> a, Vector<Integer> b)
{
// Vector to store the output
Vector<Integer> result = new Vector<Integer>();
int m = a.size(), n = b.size();
// Loop till a or b runs out
int i = 0;
while (i < m && i < n)
{
// Get sum of next element from each array
int sum = a.get(i) + b.get(i);
// Separate the digits of sum and add them to
// the resultant vector
split_number(sum, result);
i++;
}
// Process remaining elements
// of first vector, if any
while (i < m)
{
split_number(a.get(i++), result);
}
// Process remaining elements
// of second vector, if any
while (i < n)
{
split_number(b.get(i++), result);
}
// Print the resultant vector
printVector(result);
}
// Driver code
public static void main(String[] args)
{
// input vectors
int[] arr1 = {23, 5, 2, 7, 87};
Vector<Integer> a = new Vector<>();
for(Integer i:arr1)
a.add(i);
int[] arr2 = {4, 67, 2, 8};
Vector<Integer> b = new Vector<Integer>();
for(Integer i:arr2)
b.add(i);
add(a, b);
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 implementation of the # above approach # Utility function to print the # contents of the list def printVector(result): for i in result: print(i, end = " ") # Recursive function to separate the # digits of a positive integer and # add them to the given list def split_number(num, result): if num > 0: split_number(num // 10, result) result.append(num % 10) # Function to add two lists def add(a, b): # List to store the output result = [] m, n = len(a), len(b) # Loop till a or b runs out i = 0 while i < m and i < n: # Get sum of next element from # each array sum = a[i] + b[i] # Separate the digits of sum and # add them to the resultant list split_number(sum, result) i += 1 # Process remaining elements of # first list, if any while i < m: split_number(a[i], result) i += 1 # Process remaining elements of # second list, if any while i < n: split_number(b[i], result) i += 1 # Print the resultant list printVector(result) # Driver Code if __name__ == "__main__": # input lists a = [23, 5, 2, 7, 87] b = [4, 67, 2, 8] add(a, b) # This code is contributed by rituraj_jain
C#
// C# implementation of the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Utility function to print
// the contents of the vector
static void printVector(List<int> result)
{
foreach (int i in result)
{
Console.Write(i + " ");
}
}
// Recursive function to separate
// the digits of a positive integer
// and add them to the given vector
static void split_number(int num, List<int> result)
{
if (num > 0)
{
split_number(num / 10, result);
result.Add(num % 10);
}
}
// Function to add two arrays
static void add(List<int> a, List<int> b)
{
// Vector to store the output
List<int> result = new List<int>();
int m = a.Count, n = b.Count;
// Loop till a or b runs out
int i = 0;
while (i < m && i < n)
{
// Get sum of next element from each array
int sum = a[i] + b[i];
// Separate the digits of sum and add them to
// the resultant vector
split_number(sum, result);
i++;
}
// Process remaining elements
// of first vector, if any
while (i < m)
{
split_number(a[i++], result);
}
// Process remaining elements
// of second vector, if any
while (i < n)
{
split_number(b[i++], result);
}
// Print the resultant vector
printVector(result);
}
// Driver code
public static void Main(String[] args)
{
// input vectors
int[] arr1 = {23, 5, 2, 7, 87};
List<int> a = new List<int>();
foreach(int i in arr1)
a.Add(i);
int[] arr2 = {4, 67, 2, 8};
List<int> b = new List<int>();
foreach(int i in arr2)
b.Add(i);
add(a, b);
}
}
// This code is contributed by princiraj1992
Javascript
<script>
// Javascript implementation of the approach
// Utility function to print the contents of the vector
function printVector(result)
{
for (let i = 0; i < result.length; i++)
document.write(result[i] + " ");
}
// Recursive function to separate the digits of a positive
// integer and add them to the given vector
function split_number(num, result)
{
if (num > 0) {
split_number(parseInt(num / 10), result);
result.push(num % 10);
}
}
// Function to add two arrays
function add(a, b)
{
// Vector to store the output
let result = [];
let m = a.length, n = b.length;
// Loop till a or b runs out
let i = 0;
while (i < m && i < n) {
// Get sum of next element from each array
let sum = a[i] + b[i];
// Separate the digits of sum and add them to
// the resultant vector
split_number(sum, result);
i++;
}
// Process remaining elements of first vector, if any
while (i < m) {
split_number(a[i++], result);
}
// Process remaining elements of second vector, if any
while (i < n) {
split_number(b[i++], result);
}
// Print the resultant vector
printVector(result);
}
// Driver code
// input vectors
let a = [ 23, 5, 2, 7, 87 ];
let b = [ 4, 67, 2, 8 ];
add(a, b);
// This code is contributed by souravmahato348.
</script>
Producción:
2 7 7 2 4 1 5 8 7
Complejidad del tiempo: O(max(m,n))
Publicación traducida automáticamente
Artículo escrito por gfg_sal_gfg y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA