Leetcode: Odd Even Linked List

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.

Given 1->2->3->4->5->NULL,
return 1->3->5->2->4->NULL.


  • 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.

## Blog link: 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

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