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 ...

先数一下要移多少次。

移动的时候跟删除一个节点的操作一样，不过删除的元素要跟到链表的末尾。

public static ListNode oddEvenList(ListNode head)
	{
		if(head==null||head.next==null||head.next.next==null)
			return head;
		
		ListNode tail=head;
		int cnt=1;
		while(tail.next!=null)
			{
				tail=tail.next;
				cnt++;
			}
		
		ListNode curtail=tail;
		ListNode pre=head;
		ListNode p=null;
		ListNode pnext=null;
		int count=0;
		while(count<cnt/2)
		{
			count++;
			p=pre.next;
			pnext=p.next;
			pre.next=pnext;
			p.next=null;
			curtail.next=p;
			curtail=p;
			pre=pnext;
		
		}
		
		return head;
	}