通过给定数组的每个元素乘以 K 生成一个数组
给定一个大小为 N 的数组 arr[] 和一个整数 K 。任务是将数组的每个元素乘以 K 。
示例:
输入: arr[] = { 3，4 }，K = 2 输出: 6 8 说明:元素变为 32 = 6，42 = 8。
输入: arr[] = { 0，1，2 }，K = 7 输出: { 0，7，14 }
方法:给定的问题可以通过以下步骤解决:
- 遍历列表中的所有元素
- 将每个元素乘以 K
- 返回修改后的列表
下面是上述方法的实现。
C++
// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to multiply all
// the elements of array by K
void multiplyAllByK(vector<int> arr, int K)
{
int N = arr.size();
// Loop to multiply all
// the array elements
for (int i = 0; i < N; i++) {
int x = arr[i];
arr[i] = K * x;
}
// Print the modified array
for (int i = 0; i < N; i++)
cout << (arr[i]) << " ";
}
// Driver code
int main()
{
vector<int> arr;
arr.push_back(3);
arr.push_back(4);
int K = 2;
multiplyAllByK(arr, K);
return 0;
}
// This code is contributed by lokeshpotta20.
Java 语言(一种计算机语言，尤用于创建网站)
// Java code to implement above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to multiply all
// the elements of array by K
public static void multiplyAllByK(
ArrayList<Integer> arr, int K)
{
int N = arr.size();
// Loop to multiply all
// the array elements
for (int i = 0; i < N; i++) {
int x = arr.get(i);
arr.set(i, K * x);
}
// Print the modified array
for (int i = 0; i < N; i++)
System.out.print(arr.get(i) + " ");
}
// Driver code
public static void main(String[] args)
{
ArrayList<Integer> arr = new ArrayList<Integer>();
arr.add(3);
arr.add(4);
int K = 2;
multiplyAllByK(arr, K);
}
}
计算机编程语言
# Python code to implement above approach
# Function to multiply all
# the elements of array by K
def multiplyAllByK(arr, K):
n = len(arr)
for i in range(n):
x = arr[i]
arr[i] = K * x
for i in range(n):
print(arr[i], end = ' ')
# Driver code
arr = [3, 4]
K = 2
multiplyAllByK(arr, K)
# This code is contributed by Samim Hossain Mondal.
C
// C# code to implement above approach
using System;
using System.Collections.Generic;
public class GFG {
// Function to multiply all
// the elements of array by K
public static void multiplyAllByK(
List<int> arr, int K)
{
int N = arr.Count;
// Loop to multiply all
// the array elements
for (int i = 0; i < N; i++) {
int x = arr[i];
arr[i] =( K * x);
}
// Print the modified array
for (int i = 0; i < N; i++)
Console.Write(arr[i] + " ");
}
// Driver code
public static void Main(String[] args)
{
List<int> arr = new List<int>();
arr.Add(3);
arr.Add(4);
int K = 2;
multiplyAllByK(arr, K);
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// JavaScript code for the above approach
// Function to multiply all
// the elements of array by K
const multiplyAllByK = (arr, K) => {
let N = arr.length;
// Loop to multiply all
// the array elements
for (let i = 0; i < N; i++) {
let x = arr[i];
arr[i] = K * x;
}
// Print the modified array
for (let i = 0; i < N; i++)
document.write(`${arr[i]} `);
}
// Driver code
let arr = [];
arr.push(3);
arr.push(4);
let K = 2;
multiplyAllByK(arr, K);
// This code is contributed by rakeshsahni
</script>
Output
6 8
时间复杂度:O(N) T3】辅助空间: O(1)
使用λ表达式的方法:这也可以使用λ表达式来实现。
n - > n * K 其中 n 可以是特定的元素，也可以是完整的数组。
下面是上述方法的实现:
Java 语言(一种计算机语言，尤用于创建网站)
// Java code to implement above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to multiply all
// the elements of array by K
public static void multiplyAllByK(
ArrayList<Integer> arr, int K)
{
arr.replaceAll(n -> n * K);
for (Integer x : arr)
System.out.print(x + " ");
}
// Driver code
public static void main(String[] args)
{
ArrayList<Integer> arr
= new ArrayList<Integer>();
arr.add(3);
arr.add(4);
int K = 2;
multiplyAllByK(arr, K);
}
}
Python 3
# Python3 code to implement above approach
# Function to multiply all
# the elements of array by K
def multiplyAllByK(arr, K):
lambda_func = lambda n: n * K
for i in range(len(arr)):
print(lambda_func(arr[i]), end = ' ')
# Driver code
arr = [3, 4]
K = 2
multiplyAllByK(arr, K)
# This code is contributed by Samim Hossain Mondal.
C
// C# code to implement above approach
using System;
using System.Collections.Generic;
using System.Linq;
public class GFG {
// Function to multiply all
// the elements of array by K
public static void multiplyAllByK(
List<int> arr, int K)
{
var temp = arr.Select(n => n * K);
foreach (int x in temp)
Console.Write(x + " ");
}
// Driver code
public static void Main(String[] args)
{
List<int> arr
= new List<int>();
arr.Add(3);
arr.Add(4);
int K = 2;
multiplyAllByK(arr, K);
}
}
// This code is contributed by shikhasingrajput
java 描述语言
<script>
// Javascript code to implement above approach
// Function to multiply all
// the elements of array by K
function multiplyAllByK(arr, K) {
arr = arr.map(k => {
return k * K
});
for (x of arr)
document.write(x + " ");
}
// Driver code
let arr = [];
arr.push(3);
arr.push(4);
let K = 2;
multiplyAllByK(arr, K);
// This code is contributed by saurabh_jaiswal.
</script>
Output
6 8
时间复杂度:O(N) T3】辅助空间: O(1)
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