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 …

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## https://code.dennyzhang.com/odd-even-linked-list # Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def oddEvenList(self, head): """ :type head: ListNode :rtype: ListNode """ ## Idea: 2pointer: odd_tail, even_tail ## move the next element of even_tail to odd_tail ## move next for both tails ## Complexity ## Sample Data: ## 1 -> 2 -> 3 -> 4 -> 5 ->NULL ## odd_tail even_tail ## if head is None or head.next is None: return head odd_tail = head even_tail = head.next while even_tail and even_tail.next: p = even_tail.next even_tail.next = p.next p.next = odd_tail.next odd_tail.next = p # move next even_tail = even_tail.next odd_tail = odd_tail.next return head