打印通过在每个索引后反转给定 Array 形成的 Array

给定一个数组arr[]，任务是打印通过在打印每个元素后翻转整个数组从第一个索引到最后一个索引遍历给定数组形成的数组。

例子：

Input: arr = {0, 1, 2, 3, 4, 5} 
Output: 0 4 2 2 4 0
Explanation: On 1st iteration element on index 0 -> 0 is printed then the whole array is flipped:  {0, 1, 2, 3, 4, 5} -> {5, 4, 3, 2, 1, 0} 
                      On 2nd iteration element on index 1 -> 4 is printed then the whole array is flipped: : {5, 4, 3, 2, 1, 0} -> {0, 1, 2, 3, 4, 5} 
                      On 3rd iteration element on index 2 -> 2 is printed then the whole array is flipped:  {0, 1, 2, 3, 4, 5} -> {5, 4, 3, 2, 1, 0} 
                      On 2nd iteration element on index 3 -> 2 is printed then the whole array is flipped: : {5, 4, 3, 2, 1, 0} -> {0, 1, 2, 3, 4, 5} 
                      On 2nd iteration element on index 4 -> 4 is printed then the whole array is flipped:  {0, 1, 2, 3, 4, 5} -> {5, 4, 3, 2, 1, 0} 
                      On 2nd iteration element on index 5 -> 0 is printed then the whole array is flipped: : {5, 4, 3, 2, 1, 0} -> {0, 1, 2, 3, 4, 5} 

Input: arr = {0, 1, 2, 3, 4}
Output: 0 3 2 1 4

方法：给定的问题可以通过使用两指针技术来解决。这个想法是从第一个索引开始从左到右迭代数组，从倒数第二个索引开始从右到左迭代数组。

可以按照以下步骤解决问题：

• 使用指针一从左到右遍历数组，使用指针二从右到左遍历数组
• 同时打印两个指针指向的元素，并将指针递增 2 并将指针递减 2

下面是上述方法的实现：

C++

// C++ code for the above approach
#include 
using namespace std;
 
// Function to print array elements at
// every index by flipping whole array
// after printing every element
void printFlip(vector arr)
{
 
    // Initialize length of the array
    int N = arr.size();
 
    // Initialize both the pointers
    int p1 = 0, p2 = N - 2;
 
    // Iterate until both pointers
    // are not out of bounds
    while (p1 < N || p2 >= 0) {
 
        // Print the elements
        cout << arr[p1] << " ";
        if (p2 > 0)
            cout << arr[p2] << " ";
 
        // Increment p1 by 2
        p1 += 2;
 
        // Decrement p2 by 2
        p2 -= 2;
    }
}
 
// Driver code
int main()
{
 
    // Initialize the array
    vector arr = { 0, 1, 2, 3, 4 };
 
    // Call the function
    // and print the array
    printFlip(arr);
    return 0;
}
 
// This code is contributed by Potta Lokesh.

Java

// Java implementation for the above approach
 
import java.io.*;
import java.util.*;
 
class GFG {
 
    // Function to print array elements at
    // every index by flipping whole array
    // after printing every element
    public static void printFlip(int[] arr)
    {
 
        // Initialize length of the array
        int N = arr.length;
 
        // Initialize both the pointers
        int p1 = 0, p2 = N - 2;
 
        // Iterate until both pointers
        // are not out of bounds
        while (p1 < N || p2 >= 0) {
 
            // Print the elements
            System.out.print(arr[p1] + " ");
            if (p2 > 0)
                System.out.print(arr[p2] + " ");
 
            // Increment p1 by 2
            p1 += 2;
 
            // Decrement p2 by 2
            p2 -= 2;
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        // Initialize the array
        int[] arr = { 0, 1, 2, 3, 4 };
 
        // Call the function
        // and print the array
        printFlip(arr);
    }
}

Python3

# Python code for the above approach
 
# Function to print array elements at
# every index by flipping whole array
# after printing every element
def printFlip(arr):
 
    # Initialize length of the array
    N = len(arr);
 
    # Initialize both the pointers
    p1 = 0
    p2 = N - 2;
 
    # Iterate until both pointers
    # are not out of bounds
    while (p1 < N or p2 >= 0):
 
        # Print the elements
        print(arr[p1], end=" ");
        if (p2 > 0):
            print(arr[p2], end=" ");
 
        # Increment p1 by 2
        p1 += 2;
 
        # Decrement p2 by 2
        p2 -= 2;
     
# Driver Code
# Initialize the array
arr = [ 0, 1, 2, 3, 4 ];
 
# Call the function
# and print the array
printFlip(arr);
 
# This code is contributed by gfgking.

C#

// C# implementation for the above approach
using System;
 
class GFG {
 
    // Function to print array elements at
    // every index by flipping whole array
    // after printing every element
    static void printFlip(int []arr)
    {
 
        // Initialize length of the array
        int N = arr.Length;
 
        // Initialize both the pointers
        int p1 = 0, p2 = N - 2;
 
        // Iterate until both pointers
        // are not out of bounds
        while (p1 < N || p2 >= 0) {
 
            // Print the elements
            Console.Write(arr[p1] + " ");
            if (p2 > 0)
                Console.Write(arr[p2] + " ");
 
            // Increment p1 by 2
            p1 += 2;
 
            // Decrement p2 by 2
            p2 -= 2;
        }
    }
 
    // Driver code
    public static void Main()
    {
 
        // Initialize the array
        int []arr = { 0, 1, 2, 3, 4 };
 
        // Call the function
        // and print the array
        printFlip(arr);
    }
}
// This code is contributed by Samim Hossain Mondal.

Javascript

// Javascript code for the above approach
 
// Function to print array elements at
// every index by flipping whole array
// after printing every element
function printFlip(arr)
{
 
    // Initialize length of the array
    let N = arr.length;
 
    // Initialize both the pointers
    let p1 = 0, p2 = N - 2;
 
    // Iterate until both pointers
    // are not out of bounds
    while (p1 < N || p2 >= 0) {
 
        // Print the elements
        document.write(arr[p1] + " ");
        if (p2 > 0)
            document.write(arr[p2] + " ");
 
        // Increment p1 by 2
        p1 += 2;
 
        // Decrement p2 by 2
        p2 -= 2;
    }
}
 
// Driver Code
// Initialize the array
let arr = [ 0, 1, 2, 3, 4 ];
 
// Call the function
// and print the array
printFlip(arr);
 
// This code is contributed by Samim Hossain Mondal.

输出

0 3 2 1 4 

时间复杂度： O(N)
辅助空间： O(1)