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.
想的是一个一个调换， 结果发现需要用到的变量太多了，
把奇偶结点分为两个链表最后合在一起比较简单。
