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:

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

用当前链表生成两个链表, 分别由当前链表的奇数项和偶数项组成,然后收尾相连形成新链表
注意遍历链表的退出条件应该是偶数项 even!=null && even.next!=null

复杂度

时间O(n) 空间O(1)

class Solution {
    public ListNode oddEvenList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode rightHead = head.next;
        ListNode odd = head;
        ListNode even = rightHead;
        while (even!= null && even.next!= null) {
            odd.next = odd.next.next;
            odd = odd.next;
            even.next = even.next.next;
            even = even.next;
        }
        odd.next = rightHead;
        return head;
    }
}