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[12/6] 328. Odd Even Linked List

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328. 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
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# 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]:
        if not head or not head.next: return head

        curr = head
        even = curr.next
        even_start = even
        odd = curr
        curr = curr.next.next
        while curr and curr.next:
            odd.next = curr
            curr = curr.next
            even.next = curr
            curr = curr.next
            odd = odd.next
            even = even.next
        
        if curr:
            odd.next = curr
            odd = odd.next
        
        odd.next = even_start
        even.next = None

        return head
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