leetcode oddevenlinkedlist

import utils.ListNode;

public class  {
  
   * 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 1:
   *
   * Input: 1->2->3->4->5->NULL
   * Output: 1->3->5->2->4->NULL
   * Example 2:
   *
   * Input: 2->1->3->5->6->4->7->NULL
   * Output: 2->3->6->7->1->5->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 ...
   */
  public ListNode oddEvenList(ListNode head) {
    if (head == null || head.next == null) return head;
    ListNode dump = new ListNode(0);
    dump.next = head;
    ListNode odd = head;
    ListNode even = head.next;
    ListNode p = even.next;
    while(p!=null){
      //insert(odd.next, p);
      even.next = p.next;
      ListNode temp = odd.next;
      odd.next = p;
      p.next = temp;

      odd = odd.next;
      even = even.next;
      if (even == null) break;
      p = even.next;
    }
    return dump.next;
  }
  public ListNode oddEvenListII(ListNode head) {
    if (head == null || head.next == null) return head;
    ListNode headOdd = head;
    ListNode headEven = head.next;
    ListNode tailOdd = head;
    ListNode taillEven = head.next;
    while (taillEven.next != null) {
      ListNode t = taillEven.next;
      tailOdd.next = t;
      taillEven.next = t.next;
      t.next = headEven;
      taillEven = taillEven.next;
      tailOdd = tailOdd.next;
    }
    return headOdd;
  }
}