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

简单的链表操作。将原有链表按照奇数和偶数的位置来进行重新排列。

需要注意边界条件，如链表尾部是空该怎么办。

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# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution(object):
    def oddEvenList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        first_even = None
        if (head):
            first_even = head.next
        else:
            return first_even
        cur = head
        odd = True
        while (cur):
            original_next = cur.next
            if (original_next):
                if (original_next.next):
                    cur.next = original_next.next
                else:
                    if (odd):
                        cur.next = first_even
                    else:
                        cur.next = None
            else:
                if (odd):
                    cur.next = first_even
            cur = original_next
            odd = not odd
        return head