328. 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 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

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

The solution is very intuitive. But it is not trivial to write a concise and bug-free code.

A well-formed LinkedList need two pointers head and tail to support operations at both ends. The variables head and odd are the head pointer and tail pointer of one LinkedList we call oddList; the variables evenHead and even are the head pointer and tail pointer of another LinkedList we call evenList. The algorithm traverses the original LinkedList and put the odd nodes into the oddList and the even nodes into the evenList. To traverse a LinkedList we need at least one pointer as an iterator for the current node. But here the pointers odd and even not only serve as the tail pointers but also act as the iterators of the original list.

The best way of solving any linked list problem is to visualize it either in your mind or on a piece of paper. An illustration of our algorithm is following: