## DEV Community

Aroup Goldar Dhruba

Posted on

# LeetCode: Odd Even Linked List

### Problem Statement

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

Example 1:

``````Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL
``````

Example 2:

``````Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->4->NULL
``````

Note:

• The relative order inside both the even and odd groups should remain as it was in the input.
• The first node is considered odd, the second node even and so on ...

### Solution Thought Process

For this problem, we have to create two linked list:

• `OddLinkedList` - for containing odd index elements
• `EvenLinkedList` - for containing even index elements

We iterate over the linked list, we initialize a counter to 1. Whenever -

• The counter is divisible by 2, we add the node to the `OddLinkedList`
• The counter is divisible by 1, we add the node to the `EvenLinkedList`

After that, we merge the tail of `OddLinkedList` to the `EvenLinkedList` head. And in the end, we make the tail of `EvenLinkedList` point to `NULL` to make sure that the linked list has ended in `NULL`.

### Solution

``````/**
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode() : val(0), next(nullptr) {}
*     ListNode(int x) : val(x), next(nullptr) {}
*     ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
int counter = 1;
while(iterator)
{
if(counter % 2)
{
oddIterator->next = iterator;
iterator = iterator->next;
oddIterator = oddIterator->next;
oddIterator->next = NULL;
}
else {
evenIterator->next = iterator;
iterator = iterator->next;
evenIterator = evenIterator->next;
evenIterator->next = NULL;
}
counter++;
}