Odd Even Linked List

Problem

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:
Given 1->2->3->4->5->NULL,
return 1->3->5->2->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

Two Pointers Solution

public class Solution {
    public ListNode oddEvenList(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode odd = head, even = head.next, evenHead = head.next;
        //no need to check odd cause it always comes before than even
        while (even != null && even.next != null) {
            odd.next = even.next;
            odd = odd.next;
            even.next = odd.next;
            even = even.next;
        }
        odd.next = evenHead;
        return head;
    }
}

Analysis

A simple solution using two pointers pointing to odd and even respectively
We need to have one extra var head.next as the evenHead
When the loop ends, we connect odd with the evenHead