题目：Odd Even Linked List

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 ...

题意：将链表按照位置的奇偶重新排序，使奇数位置的元素按照顺序在前，然后在接偶数位置的元素。

例如：1->2->3->4->5->NULL，重排后1->3->5->2->4->NULL.

思路：简单来先将链表分为奇偶两个链表在连接两个链表，注意啊要保存链表的首尾。

ListNode* oddEvenList(ListNode* head) {
        if (!head || !(head->next))return head;
        ListNode *odd = head, *even = head->next;
        ListNode *p = even->next, *op = odd, *ep = even;
        while (p){
            op->next = p;
            op = p;
            p = p->next;
            if (!p)break;
            ep->next = p;
            ep = p;
            p = p->next;
        }
        op->next = even;
        ep->next = nullptr;
        return odd;
    }

[LeetCode]Odd Even Linked List