# Reversed Linked List

Problem:-

You are given a linked list that contains

$N$

integers. You have performed the following reverse operation on the list:

• Select all the subparts of the list that contain only even integers. For example, if the list is
$\left\{1,2,8,9,12,16\right\}$

, then the selected subparts will be

$\left\{2,8\right\}$

$\left\{12,16\right\}$

.

• Reverse the selected subpart such as
$\left\{8,2\right\}$

and

$\left\{16,12\right\}$

.

Now, you are required to retrieve the original list.

Note: You should use the following definition of the linked list for this problem:

``class Node {    Object data;    Node next;}``

Input format

• First line:
$N$

• Next line:
$N$

space-separated integers that denote elements of the reverse list

Output format

Print the

$N$

elements of the original list.

Constraints

$1\le N\le {10}^{3}$

$1\le {A}_{i}\le {10}^{9}$

Sample Input
`92 18 24 3 5 7 9 6 12`
Sample Output
`24 18 2 3 5 7 9 12 6`
Time Limit: 1
Memory Limit: 256
Source Limit:
Explanation

In the sample, the original list is

$\left\{24,18,2,3,5,7,9,12,6\right\}$

which when reversed according to the operations will result in the list given in the sample input.

Solution:-

#include<stdio.h>
void main()
{
int n,j,i,k;
long int a,b;
scanf(“%d”,&n);
for(i=0;i<n;i++)
{
scanf(“%ld”,&a[i]);
}
for(i=0;i<n;i++)
{
if(a[i]%2!=0)
printf(“%ld “,a[i]);
else
{
j=0;
while(1)
{
b[j]=a[i];
j++;
i++;
if(a[i]%2!=0||i==n)
break;
}
for(k=j1;k>=0;k–)
printf(“%ld “,b[k]);
i–;
}
}
}