Odd Even Linked List（奇偶链表，把所有第偶数个节点排到所有奇数个结点后面，相对顺序不变）

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： 1、先把所有第奇数个结点连接起来，所有第偶数个结点连接起来；2、最后一个奇结点连接到第一个偶节点。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution 
{
public:
    ListNode* oddEvenList(ListNode* head) 
    {
        if(!head) return head;
        ListNode* p=head;
        ListNode* q=head->next;
        ListNode* r=head->next;
        while(p->next&&q->next)
        {
            p->next=q->next;
            p=p->next;
            q->next=p->next;
            q=q->next;
        }
        p->next=r;

        return head;
    }
};