Leetcode 328. Odd Even Linked List

题目：
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：使用两个指针，一个指针串联所有奇数位置的结点，一个指针串联所有偶数位置的结点，最后将偶数节点的指针拼接到计数结点串的后面。

代码：

public ListNode oddEvenList(ListNode head) {
        if(head == null){
            return null;
        }
        ListNode odd = head.next;
        ListNode even = head;
        ListNode phead = odd;
        if(odd == null){
            return head;
        }

        while(odd.next != null && even.next != null){
            even.next = odd.next;
            even = even.next;
            if(even.next!=null){
                odd.next = even.next;
                odd =odd.next;
            }else{
                odd.next = null;
                break;
            }
        }
        even.next = phead;
        return head;
    }