Odd Even Linked List

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:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def getList(self, head, rem, i):
        if not head:
            return head
        if i % 2 != rem:
            return self.getList(head.next, rem, i + 1)
        newhead = TreeNode(val = head.val)
        newhead.next = self.getList(head.next, rem, i + 1)
        return newhead
    
    def oddEvenList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        left = self.getList(head, 1, 1)
        right = self.getList(head, 0, 1)
        if not left:
            return right
        curr = left
        while curr and curr.next:
            curr = curr.next
        curr.next = right
        return left