328. 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 …

//注意保留一个even_head的位置，最后要连接被移动的odd的末尾到一直保留的even_head的开头。这时复制的even节点even已经在最末尾

public class Solution   
{  
    public static ListNode oddEvenList(ListNode head)   
    {  
        if (head == null || head.next == null)  
            return head;  
        ListNode even_head = head.next;  
        ListNode odd = head;
        ListNode even = even_head;  
        while(even != null && even.next != null)   
        {  
            odd.next = even.next;  
            even.next = even.next.next;  
            odd = odd.next;  
            even = even.next;  
        }  
        odd.next = even_head;  
        return head;  
    }  
}