打印阵列的替代元素
原文:https://www . geesforgeks . org/print-alternate-elements of-a-array/
给定一个大小为 N 的数组、 arr[] ，任务是打印给定数组中出现在奇数索引处的元素(基于 1 的索引)。
示例:
输入: arr[] = {1，2，3，4，5} 输出: 1 3 5 解释: 出现在奇数位置的数组元素为:{1，3，5}。 因此，需要的输出是 1 3 5。
输入: arr[] = {-5，1，4，2，12 } T3】输出: -5 4 12
天真法:解决这个问题最简单的方法是遍历给定数组，检查当前元素的位置是否为奇数。如果发现为真，则打印当前元素。
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print
// Alternate elements
// of the given array
void printAlter(int arr[], int N)
{
// Print elements
// at odd positions
for (int currIndex = 0;
currIndex < N; currIndex++) {
// If currIndex stores even index
// or odd position
if (currIndex % 2 == 0) {
cout << arr[currIndex] << " ";
}
}
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int N = sizeof(arr) / sizeof(arr[0]);
printAlter(arr, N);
}
C
// C program to implement
// the above approach
#include <stdio.h>
// Function to print
// Alternate elements
// of the given array
void printAlter(int arr[], int N)
{
// Print elements
// at odd positions
for (int currIndex = 0;
currIndex < N; currIndex++) {
// If currIndex stores even index
// or odd position
if (currIndex % 2 == 0) {
printf("%d ", arr[currIndex]);
}
}
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int N = sizeof(arr) / sizeof(arr[0]);
printAlter(arr, N);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to implement
// the above approach
import java.io.*;
class GFG{
// Function to print
// Alternate elements
// of the given array
static void printAlter(int[] arr, int N)
{
// Print elements
// at odd positions
for(int currIndex = 0;
currIndex < N;
currIndex++)
{
// If currIndex stores even index
// or odd position
if (currIndex % 2 == 0)
{
System.out.print(arr[currIndex] + " ");
}
}
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 1, 2, 3, 4, 5 };
int N = arr.length;
printAlter(arr, N);
}
}
// This code is contributed by akhilsaini
Python 3
# Python3 program to implement
# the above approach
# Function to print
# Alternate elements
# of the given array
def printAlter(arr, N):
# Print elements
# at odd positions
for currIndex in range(0, N):
# If currIndex stores even index
# or odd position
if (currIndex % 2 == 0):
print(arr[currIndex], end = " ")
# Driver Code
if __name__ == "__main__":
arr = [ 1, 2, 3, 4, 5 ]
N = len(arr)
printAlter(arr, N)
# This code is contributed by akhilsaini
C
// C# program to implement
// the above approach
using System;
class GFG{
// Function to print
// Alternate elements
// of the given array
static void printAlter(int[] arr, int N)
{
// Print elements
// at odd positions
for(int currIndex = 0;
currIndex < N;
currIndex++)
{
// If currIndex stores even index
// or odd position
if (currIndex % 2 == 0)
{
Console.Write(arr[currIndex] + " ");
}
}
}
// Driver Code
public static void Main()
{
int[] arr = { 1, 2, 3, 4, 5 };
int N = arr.Length;
printAlter(arr, N);
}
}
// This code is contributed by akhilsaini
java 描述语言
// javascript program to implement
// the above approach
// Function to print
// Alternate elements
// of the given array
function printAlter(arr, N)
{
// Print elements
// at odd positions
for(var currIndex = 0; currIndex < N; currIndex++)
{
// If currIndex stores even index
// or odd position
if (currIndex % 2 == 0)
{
document.write(arr[currIndex] + " ");
}
}
}
// Driver Code
var arr = [ 1, 2, 3, 4, 5 ]
var N = arr.length;
printAlter(arr, N);
// This code is contributed by bunnyram19.
Output:
1 3 5
时间复杂度:O(N) T5辅助空间:** O(1)
有效方法:为了优化上述方法，其思想是只遍历给定数组中出现在奇数位置的那些元素。按照以下步骤解决问题:
- 用循环变量迭代一个从 0 到 N 的循环。
- 打印arr【currIndex】的值，并将 currIndex 的值增加 2 直到 currIndex 超过 n
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print
// Alternate elements
// of the given array
void printAlter(int arr[], int N)
{
// Print elements
// at odd positions
for (int currIndex = 0;
currIndex < N; currIndex += 2) {
// Print elements of array
cout << arr[currIndex] << " ";
}
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int N = sizeof(arr) / sizeof(arr[0]);
printAlter(arr, N);
}
C
// C program to implement
// the above approach
#include <stdio.h>
// Function to print
// Alternate elements
// of the given array
void printAlter(int arr[], int N)
{
// Print elements
// at odd positions
for (int currIndex = 0;
currIndex < N; currIndex += 2) {
// Print elements of array
printf("%d ", arr[currIndex]);
}
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int N = sizeof(arr) / sizeof(arr[0]);
printAlter(arr, N);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to implement
// the above approach
import java.io.*;
class GFG{
// Function to print
// Alternate elements
// of the given array
static void printAlter(int[] arr, int N)
{
// Print elements
// at odd positions
for(int currIndex = 0;
currIndex < N;
currIndex += 2)
{
// Print elements of array
System.out.print(arr[currIndex] + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 1, 2, 3, 4, 5 };
int N = arr.length;
printAlter(arr, N);
}
}
// This code is contributed by akhilsaini
Python 3
# Python3 program to implement
# the above approach
# Function to print
# Alternate elements
# of the given array
def printAlter(arr, N):
# Print elements
# at odd positions
for currIndex in range(0, N, 2):
# Print elements of array
print(arr[currIndex], end = " ")
# Driver Code
if __name__ == "__main__":
arr = [ 1, 2, 3, 4, 5 ]
N = len(arr)
printAlter(arr, N)
# This code is contributed by akhilsaini
C
// C# program to implement
// the above approach
using System;
class GFG{
// Function to print
// Alternate elements
// of the given array
static void printAlter(int[] arr, int N)
{
// Print elements
// at odd positions
for(int currIndex = 0;
currIndex < N;
currIndex += 2)
{
// Print elements of array
Console.Write(arr[currIndex] + " ");
}
}
// Driver Code
public static void Main()
{
int[] arr = { 1, 2, 3, 4, 5 };
int N = arr.Length;
printAlter(arr, N);
}
}
// This code is contributed by akhilsaini
java 描述语言
<script>
// Function to print
// Alternate elements
// of the given array
function printAlter(arr, N)
{
// Print elements
// at odd positions
for (var currIndex = 0;
currIndex < N; currIndex += 2)
{
// Print elements of array
document.write( arr[currIndex] + " ");
}
}
var arr= [ 1, 2, 3, 4, 5 ];
var N = 5;
printAlter(arr, N);
</script>
Output:
1 3 5
时间复杂度:O(N) T5辅助空间:** O(1)
方法 3:在 Python 中使用切片:
通过设置步长值为 2，使用 python 列表切片 进行切片。
下面是实现:
Python 3
# Python3 program to implement
# the above approach
# Function to print
# Alternate elements
# of the given array
def printAlter(arr, N):
# Print elements
# at odd positions by using slicing
# we use * to print with spaces
print(*arr[::2])
# Driver Code
if __name__ == "__main__":
arr = [1, 2, 3, 4, 5]
N = len(arr)
printAlter(arr, N)
# This code is contributed by vikkycirus
输出:
1 3 5
时间复杂度: O(N)
空间复杂度: O(1)
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