添加具有给定约束的给定数组的元素
给定两个整数数组，通过满足以下约束将它们的元素添加到第三个数组中 -
1. 加法应该从两个数组的第 0 个索引开始。
2. 如果总和不是单个数字，则拆分总和，并将数字存储在输出数组中的相邻位置。
3. 输出数组应容纳较大输入数组的任何剩余数字。
例子：
Input:
a = [9, 2, 3, 7, 9, 6]
b = [3, 1, 4, 7, 8, 7, 6, 9]
Output:
[1, 2, 3, 7, 1, 4, 1, 7, 1, 3, 6, 9]
Input:
a = [9343, 2, 3, 7, 9, 6]
b = [34, 11, 4, 7, 8, 7, 6, 99]
Output:
[9, 3, 7, 7, 1, 3, 7, 1, 4, 1, 7, 1, 3, 6, 9, 9]
Input:
a = []
b = [11, 2, 3 ]
Output:
[1, 1, 2, 3 ]
Input:
a = [9, 8, 7, 6, 5, 4, 3, 2, 1]
b = [1, 2, 3, 4, 5, 6, 7, 8, 9]
Output:
[1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0]难度级别：新手
这个想法很简单。我们维护一个输出数组并从两个数组的第 0 个索引处运行一个循环。对于循环的每次迭代，我们都会考虑两个数组中的下一个元素并添加它们。如果总和大于 9，我们将总和的各个数字推送到输出数组，否则我们推送总和本身。最后，我们将较大输入数组的剩余元素推送到输出数组。
下面是上述想法的实现：
C++
// C++ program to add two arrays following given
// constraints
#include
using namespace std;
// Function to push individual digits of a number
// to output vector from left to right
void split(int num, vector &out)
{
vector arr;
while (num)
{
arr.push_back(num%10);
num = num/10;
}
// reverse the vector arr and append it to output vector
out.insert(out.end(), arr.rbegin(), arr.rend());
}
// Function to add two arrays keeping given
// constraints
void addArrays(int arr1[], int arr2[], int m, int n)
{
// create a vector to store output
vector out;
// maintain a variable to store current index in
// both arrays
int i = 0;
// loop till arr1 or arr2 runs out
while (i < m && i < n)
{
// read next elements from both arrays and
// add them
int sum = arr1[i] + arr2[i];
// if sum is single digit number
if (sum < 10)
out.push_back(sum);
else
{
// if sum is not a single digit number, push
// individual digits to output vector
split(sum, out);
}
// increment to next index
i++;
}
// push remaining elements of first input array
// (if any) to output vector
while (i < m)
split(arr1[i++], out);
// push remaining elements of second input array
// (if any) to output vector
while (i < n)
split(arr2[i++], out);
// print the output vector
for (int x : out)
cout << x << " ";
}
// Driver code
int main()
{
int arr1[] = {9343, 2, 3, 7, 9, 6};
int arr2[] = {34, 11, 4, 7, 8, 7, 6, 99};
int m = sizeof(arr1) / sizeof(arr1[0]);
int n = sizeof(arr2) / sizeof(arr2[0]);
addArrays(arr1, arr2, m, n);
return 0;
} Java
// Java program to add two arrays following given
// constraints
import java.util.Vector;
class GFG
{
// Function to push individual digits of a number
// to output vector from left to right
static void split(int num, Vector out)
{
Vector arr = new Vector<>();
while (num > 0)
{
arr.add(num % 10);
num /= 10;
}
// reverse the vector arr and
// append it to output vector
for (int i = arr.size() - 1; i >= 0; i--)
out.add(arr.elementAt(i));
}
// Function to add two arrays keeping given
// constraints
static void addArrays(int[] arr1, int[] arr2,
int m, int n)
{
// create a vector to store output
Vector out = new Vector<>();
// maintain a variable to store
// current index in both arrays
int i = 0;
// loop till arr1 or arr2 runs out
while (i < m && i < n)
{
// read next elements from both arrays
// and add them
int sum = arr1[i] + arr2[i];
// if sum is single digit number
if (sum < 10)
out.add(sum);
else
// if sum is not a single digit number,
// push individual digits to output vector
split(sum, out);
// increment to next index
i++;
}
// push remaining elements of first input array
// (if any) to output vector
while (i < m)
split(arr1[i++], out);
// push remaining elements of second input array
// (if any) to output vector
while (i < n)
split(arr2[i++], out);
// print the output vector
for (int x : out)
System.out.print(x + " ");
}
// Driver Code
public static void main(String[] args)
{
int[] arr1 = { 9343, 2, 3, 7, 9, 6 };
int[] arr2 = { 34, 11, 4, 7, 8, 7, 6, 99 };
int m = arr1.length;
int n = arr2.length;
addArrays(arr1, arr2, m, n);
}
}
// This code is contributed by
// sanjeev2552 C#
// C# program to add two arrays following given
// constraints
using System;
using System.Collections.Generic;
class GFG
{
// Function to push individual digits of a number
// to output vector from left to right
static void split(int num, List outs)
{
List arr = new List();
while (num > 0)
{
arr.Add(num % 10);
num /= 10;
}
// reverse the vector arr and
// append it to output vector
for (int i = arr.Count - 1; i >= 0; i--)
outs.Add(arr[i]);
}
// Function to add two arrays keeping given
// constraints
static void addArrays(int[] arr1, int[] arr2,
int m, int n)
{
// create a vector to store output
List outs = new List();
// maintain a variable to store
// current index in both arrays
int i = 0;
// loop till arr1 or arr2 runs out
while (i < m && i < n)
{
// read next elements from both arrays
// and add them
int sum = arr1[i] + arr2[i];
// if sum is single digit number
if (sum < 10)
outs.Add(sum);
else
// if sum is not a single digit number,
// push individual digits to output vector
split(sum, outs);
// increment to next index
i++;
}
// push remaining elements of first input array
// (if any) to output vector
while (i < m)
split(arr1[i++], outs);
// push remaining elements of second input array
// (if any) to output vector
while (i < n)
split(arr2[i++], outs);
// print the output vector
foreach (int x in outs)
Console.Write(x + " ");
}
// Driver Code
public static void Main(String[] args)
{
int[] arr1 = { 9343, 2, 3, 7, 9, 6 };
int[] arr2 = { 34, 11, 4, 7, 8, 7, 6, 99 };
int m = arr1.Length;
int n = arr2.Length;
addArrays(arr1, arr2, m, n);
}
}
// This code is contributed by PrinciRaj1992 Javascript
输出：
9 3 7 7 1 3 7 1 4 1 7 1 3 6 9 9上述解决方案的时间复杂度为 O(m + n)，因为我们恰好遍历两个数组一次。