328. 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 ...

05/02/2018 update:

Convert the linkedlist to two linked lists above.
There are two reasons to have this while condition.
1. when even.next == null, we reached the end.
2. odd.next points to even.next then odd goes one step further to odd.next. We have to make sure even.next is not null.
Analysis:

It's similar to copy randomlist. Just the last step split of copy random list. But have to be cautious about the while loop condition. Since even.next and even is been assigned to, so they should not be null.

Solution:

class Solution {
    public ListNode oddEvenList(ListNode head) {
        if (head == null) return null;
        ListNode odd = head, even = head.next;
        ListNode evenHead = 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;
    }
}