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:
Given1->2->3->4->5->NULL,
return1->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: 维护odd_tail, even head, even tail指针。

    public ListNode oddEvenList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode odd_tail = head, even_head = head.next, even_tail = even_head, cur = even_tail.next;
        while (cur != null) {
            even_tail.next = cur.next;
            even_tail = even_tail.next;
            odd_tail.next = cur;
            odd_tail = odd_tail.next;
            cur.next = even_head;
            //要有这，eg: 1->2->3->null
            if (even_tail == null) {
                break;
            }
            cur = even_tail.next;
        }
        return head;
    }