LeetCode 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


  • 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 …


Create an odd list and an even list, then combine together.

Python Solution

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def oddEvenList(self, head: ListNode) -> ListNode:
        dummy = ListNode(0)
        odd_dummy = ListNode(0)
        even_dummy = ListNode(0)
        dummy.next = head
        count = 1
        current = head
        odd = odd_dummy
        even = even_dummy
        while current != None:
            if count % 2 == 0:
                if count == 2:
                    even_dummy.next = current
                even.next = current
                even = even.next                
                if count == 1:
                    odd_dummy.next = current                
                odd.next = current
                odd = odd.next
            current = current.next
            count += 1

        even.next = None
        odd.next = even_dummy.next            
        dummy.next = odd_dummy.next
        return dummy.next
  • Time complexity: O(N).
  • Space complexity: O(1).

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