[Solved] 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.

Table of Contents

Question

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:

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

Example 2:

oddeven2 linked list
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

Python Solution

# 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: ListNode) -> ListNode:
        if not head:return 
        
        el = head.next
        ol = head
        
        p = el
        
        while el and el.next:
            ol.next = ol.next.next
            el.next = el.next.next
            ol = ol.next
            el = el.next
        
        ol.next = p
        return head

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