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/**

* 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 ...

*

*/

/**

* Definition for singly-linked list.

* public class ListNode {

* int val;

* ListNode next;

* ListNode(int x) { val = x; }

* }

*/

public class OddEvenLinkedList328 {

public ListNode oddEvenList(ListNode head) {

if (head == null) return null;

ListNode dummy = new ListNode(0);

dummy.next = head;

ListNode evens = new ListNode(0);

ListNode e = evens;

ListNode p = dummy;

while (p.next != null) {

p = p.next;

if (p.next == null) break;

e.next = new ListNode(p.next.val);

e = e.next;

p.next = p.next.next;

}

p.next = evens.next;

return dummy.next;

}

public ListNode oddEvenList2(ListNode head) {

if (head == null) return null;

ListNode dummy = new ListNode(0);

dummy.next = head;

ListNode evens = new ListNode(0);

ListNode e = evens;

ListNode p = dummy;

while (p.next != null) {

p = p.next;

if (p.next == null) break;

e.next = p.next;

e = e.next;

p.next = p.next.next;

e.next = null;

}

p.next = evens.next;

return dummy.next;

}

/**

* https://leetcode.com/problems/odd-even-linked-list/solution/

*/

public ListNode oddEvenList3(ListNode head) {

if (head == null) return null;

ListNode odd = head, even = head.next, evenHead = even;

while (even != null && even.next != null) {

odd.next = even.next;

odd = odd.next;

even.next = odd.next;

even = even.next;

}

odd.next = evenHead;

return head;

}

}