给定一个包含N个整数的数组arr [] ，任务是重新排列该数组，以使奇数索引元素的顺序相反。
例子：
Input: arr[] = {5, 7, 6, 2, 9, 18, 11, 15}
Output: {5, 15, 6, 18, 9, 2, 11, 7}
Explanation:
The elements at even index [5, 6, 9, 11] are unchanged and elements at odd index are reversed from [7, 2, 18, 15] to [15, 18, 2, 7].
Input: arr[] = {1, 2, 3, 4, 5, 6}
Output: {1, 6, 3, 4, 5, 2}
Explanation:
The elements at even index are unchanged and elements at odd index are reversed from [2, 4, 6] to [6, 4, 2].
方法：要解决上述问题，请执行以下步骤：
- 将给定数组的奇数索引元素推入堆栈数据结构。
- 将当前位于奇数索引处的数组元素替换为堆栈顶部的元素，并不断弹出直到堆栈为空。
下面是上述方法的实现：
C++
// C++ program to reverse the
// elements only at odd positions
// in the given Array
#include
using namespace std;
// Function to display elements
void show(int arr[], int n)
{
cout << "{";
for (int i = 0; i < n - 1; i++)
cout << arr[i] << ", ";
cout << arr[n - 1] << "}";
}
// Function to flip elements
// at odd indexes
void flipHalf(int arr[], int n)
{
int c = 0;
int dup = n;
stack st;
// Pushing elements at odd indexes
// of a array to a stack
for (int i = 0; i < n; i++) {
int x = arr[i];
if (c % 2 == 1)
st.push(x);
c++;
}
c = 0;
// Replacing current elements at odd
// indexes with element at top of stack
for (int i = 0; i < n; i++) {
int x = arr[i];
if (c % 2 == 1) {
x = st.top();
st.pop();
}
arr[i] = x;
c++;
}
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
flipHalf(arr, n);
show(arr, n);
return 0;
} Java
// Java program to reverse the
// elements only at odd positions
// in the given array
import java.io.*;
import java.util.*;
class GFG {
// Function to count the valley points
// in the given character array
static void show(int arr[], int n)
{
System.out.print("{");
for(int i = 0; i < n - 1; i++)
System.out.print(arr[i] + ", ");
System.out.print(arr[n - 1] + "}");
}
// Function to flip elements
// at odd indexes
public static void flipHalf(int arr[], int n)
{
int c = 0;
int dup = n;
Stack st = new Stack<>();
// Pushing elements at odd indexes
// of a array to a stack
for(int i = 0; i < n; i++)
{
int x = arr[i];
if (c % 2 == 1)
{
st.push(x);
}
c++;
}
c = 0;
// Replacing current elements at odd
// indexes with element at top of stack
for(int i = 0; i < n; i++)
{
int x = arr[i];
if (c % 2 == 1)
{
x = st.peek();
st.pop();
}
arr[i] = x;
c++;
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4, 5, 6 };
int n = arr.length;
flipHalf(arr, n);
show(arr, n);
}
}
// This code is contributed by Aman Kumar 27 Python3
# Python3 program to reverse the
# elements only at odd positions
# in the given Array
# Function to get top of the stack
def peek_stack(stack):
if stack:
return stack[-1]
# Function to display elements
def show(arr, n):
print("{", end = " ")
for i in range(0, n - 1):
print(arr[i], ",", end = " ")
print(arr[n - 1] , "}")
# Function to flip elements
# at odd indexes
def flipHalf(arr, n):
c = 0
dup = n
stack = []
# Pushing elements at odd indexes
# of a array to a stack
for i in range(0, n):
x = arr[i]
if c % 2 == 1:
stack.append(x)
c = c + 1
c = 0
# Replacing current elements at odd
# indexes with element at top of stack
for i in range(0, n):
x = arr[i]
if c % 2 == 1:
x = peek_stack(stack)
stack.pop()
arr[i] = x
c = c + 1
# Driver Code
if __name__ == "__main__":
arr = [ 1, 2, 3, 4, 5, 6 ]
n = len(arr)
flipHalf(arr, n)
show(arr, n)
# This code is contributed by akhilsainiC#
// C# program to reverse the
// elements only at odd positions
// in the given array
using System;
using System.Collections.Generic;
class GFG{
// Function to count the valley points
// in the given character array
static void show(int []arr, int n)
{
Console.Write("{");
for(int i = 0; i < n - 1; i++)
Console.Write(arr[i] + ", ");
Console.Write(arr[n - 1] + "}");
}
// Function to flip elements
// at odd indexes
public static void flipHalf(int []arr, int n)
{
int c = 0;
int dup = n;
Stack st = new Stack();
// Pushing elements at odd indexes
// of a array to a stack
for(int i = 0; i < n; i++)
{
int x = arr[i];
if (c % 2 == 1)
{
st.Push(x);
}
c++;
}
c = 0;
// Replacing current elements at odd
// indexes with element at top of stack
for(int i = 0; i < n; i++)
{
int x = arr[i];
if (c % 2 == 1)
{
x = st.Peek();
st.Pop();
}
arr[i] = x;
c++;
}
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 2, 3, 4, 5, 6 };
int n = arr.Length;
flipHalf(arr, n);
show(arr, n);
}
}
// This code is contributed by 29AjayKumar 输出：
{1, 6, 3, 4, 5, 2}
时间复杂度： O(N)
辅助空间复杂度： O(N / 2)