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 1:

Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL

Example 2:

Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->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 ..

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None
解法：
1.长度小于3直接返回
2.奇数位置和偶数位置分别各自成链
3.拼接2个链表
class Solution(object):
    def oddEvenList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next or not head.next.next:
            return head
        
        odd = head
        even_head= head.next
        
        while odd and odd.next and odd.next.next:
            even = odd.next
            odd.next = odd.next.next
            even.next = even.next.next
            odd = odd.next
        
        odd.next = even_head
        return head