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 ...

Credits:

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 简单的链表问题，将位置小标为奇数的节点和偶数的节点分别链接成两个链表，再将两个链表首尾相连即可。

/** * 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||!head->next)            return head;        ListNode *odd = head;        ListNode *even = head->next;        ListNode *even_head=even;        while(even&&even->next){            odd->next=even->next;            odd=odd->next;            even->next=odd->next;            even=even->next;        }        odd->next=even_head;        return head;    }};