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328. Odd Even Linked List

Difficulty Topics

Description

Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list.

The first node is considered odd, and the second node is even, and so on.

Note that the relative order inside both the even and odd groups should remain as it was in the input.

You must solve the problem in O(1) extra space complexity and O(n) time complexity.

 

Example 1:

Input: head = [1,2,3,4,5]
Output: [1,3,5,2,4]

Example 2:

Input: head = [2,1,3,5,6,4,7]
Output: [2,3,6,7,1,5,4]

 

Constraints:

  • The number of nodes in the linked list is in the range [0, 104].
  • -106 <= Node.val <= 106

Solution

odd-even-linked-list.py
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def oddEvenList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        o = odd = ListNode(-1)
        e = even = ListNode(-1)

        isOdd = True
        while head:
            node = ListNode(head.val)
            if isOdd:
                odd.next = node
                odd = odd.next
            else:
                even.next = node
                even = even.next

            isOdd = not isOdd
            head = head.next

        odd.next = e.next

        return o.next