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# odd-even-linkedlist

## Problem

[Odd Even Linked List](https://leetcode.com/explore/featured/card/may-leetcoding-challenge/536/week-3-may-15th-may-21st/3331/)

## Problem Description

```
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:

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

## Solution

This problem is intuitive, gather all odd nodes together, and all even nodes together. then odd list point to even node list.

List Node is a bit not intuitive to implement, so you need to understand pointer for list node. Will using an example below to demonstrate how pointer works and how to gather all odd and even nodes.

* will need 3 new heads,
  * odd -- point to odd nodes, while iterate through list nodes.
  * even -- point to even nodes
  * evenHead -- need to remember the start position for even head.
* init, `odd = head` and `even = head.next` and `evenHead = even`
* After init, traverse through list nodes, here head and evenHead  will remain, not move, instead, odd and even will move to corresponding position. 
* each traverse, `odd = odd.next.next` (by pass even node), and `even = even.next.next` (by passing odd node)
* until `even == null || even.next == null`, done scanning all nodes, evenHead is in the even list nodes head.
* point `odd.next = evenHead`.
* now head is the result.

For example:

![Odd Even Linked List](/files/-MKvgUxKbZdaHW2lYUIZ)

### Complexity Analysis

**Time Complexity:** `O(N)`

**Space Complexity:** `O(1)`

* N - the number of nodes in linked list 

### Code

```java
class Solution {
    /**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode oddEvenList(ListNode head) {
        if (head == null) return head;
        ListNode odd = head;
        ListNode even = head.next;
        ListNode evenHead = even;
        while (even != null && even.next != null) {
            odd.next = odd.next.next;
            even.next = even.next.next;
            odd = odd.next;
            even = even.next;
        }
        odd.next = evenHead;
        return head;
    }
}
}
```