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 ...

思路：用两个oddhead和evenhead去hold两个list，然后cur往后走，遇见odd加入odd list，遇见even加入even list；

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode oddEvenList(ListNode head) {
        if(head == null || head.next == null) {
            return head;
        }
        ListNode oddhead = new ListNode(0);
        ListNode oddcur = oddhead;
        ListNode evenhead = new ListNode(0);
        ListNode evencur = evenhead;

        ListNode cur = head;
        boolean isodd = true;
        while(cur != null) {
            ListNode curnext = cur.next;
            cur.next = null;
            if(isodd) {
                oddcur.next = cur;
                oddcur = cur;
            } else {
                evencur.next = cur;
                evencur = cur;
            }
            isodd = !isodd;
            cur = curnext;
        }
        oddcur.next = evenhead.next;
        return oddhead.next;
    }
}