Odd Even Linked List

Problem

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 ...

Solution

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode oddEvenList(ListNode head) {
        if(head == null || head.next == null) return head;

        ListNode evenHead = head.next, odd = head, even = evenHead;
        while(even != null && even.next != null) {
            odd.next = even.next;
            even.next = odd.next.next; // No need to check null cause even.next != null,
            odd = odd.next;
            even = even.next;
        }
        odd.next = evenHead;
        return head;
    }
}

Analysis

This question requires O(1) extra space and O(#nodes) time complexity.
At first, I thought creating three nodes before the loop is O(3) space.
However, they all only count O(1) space cause they are not like map or array.
The key to this question is just connecting the link in correct positions, pretty straightforward.