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 ...

思路：奇数node用个dummy node连接起来，偶数的node用个dummy node连接起来。最后再拼接起来。注意要把偶数node最后接上null node。O(n)

    public ListNode oddEvenList(ListNode head) {        ListNode odd=new ListNode(-1);        ListNode copyodd=odd;        ListNode even=new ListNode(-1);        ListNode copyeven=even;        int i=1;        while(head!=null)        {            if(i%2!=0)            {                copyodd.next=head;                copyodd=copyodd.next;            }            else            {                copyeven.next=head;                copyeven=copyeven.next;            }            head=head.next;            i++;        }        copyeven.next=null;        copyodd.next=even.next;        return odd.next;            }

328. Odd Even Linked List