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 ...

奇数位置的listnode放到偶数位前面。

思路：1.奇数一个链表，偶数一个链表

         2.奇数尾巴接偶数头（所有奇数尾巴一定不是None，提前判断，偶数尾巴一定要是None，防止环）。

代码如下：

class Solution(object):
    def oddEvenList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next:
            return head
        first = head
        second = head.next
        curEven = second
        while (first.next and first.next.next) or (curEven.next and curEven.next.next):
            if first.next and first.next.next:
                first.next = first.next.next
                first = first.next
            if curEven.next and curEven.next.next:
                curEven.next = curEven.next.next
                curEven = curEven.next
        curEven.next = None
        first.next = second
        return head

简洁版本，变慢了，囧：

class Solution(object):
    def oddEvenList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next:
            return head
        first = head
        second = head.next
        curEven = second
        while curEven and curEven.next:
                first.next = first.next.next
                curEven.next = curEven.next.next
                first = first.next
                curEven = curEven.next
        first.next = second
        return head