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

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 ...

題目：將原連結串列奇數位移動到偶數位前，（注意不是節點值奇偶）原來奇偶序列中順序不變。

思路：這個在迴圈中每次移動兩位，第一位為奇數位，第二位為偶數位，分別放進奇偶連結串列指標末尾。

程式碼：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode oddEvenList(ListNode head) {
        if(head==null||head.next==null||head.next.next==null)
            return head;
        ListNode odd = head;
        ListNode even =head.next;
        ListNode cur = even;
        //head=head.next;
        ListNode node = head.next.next; 
        while(node!=null){
            odd.next=node;
            node = node.next;
            odd = odd.next;
            if(node==null)
                break;
            cur.next=node;
            node = node.next;
            cur=cur.next;
        }
        odd.next=even;
        cur.next=null;
        return head;
    }
}